Does the series $sum_{n=1}^infty frac{(-1)^{n+1}}{3n+n(-1)^n}$ converge?












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I have to find out if the series $$sum_{n=1}^infty frac{(-1)^{n+1}}{3n+n(-1)^n}$$ converges. Root test and ratio test did not work out for me. I also tried the alternating series test, but I can not use it because of the $${(-1)^{n+1}}$$ in the denominator.
Also, I have to find the limit of the series if it does converge.






sequences-and-series limits convergence divergent-series







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    I have to find out if the series $$sum_{n=1}^infty frac{(-1)^{n+1}}{3n+n(-1)^n}$$ converges. Root test and ratio test did not work out for me. I also tried the alternating series test, but I can not use it because of the $${(-1)^{n+1}}$$ in the denominator.
    Also, I have to find the limit of the series if it does converge.






    sequences-and-series limits convergence divergent-series







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      I have to find out if the series $$sum_{n=1}^infty frac{(-1)^{n+1}}{3n+n(-1)^n}$$ converges. Root test and ratio test did not work out for me. I also tried the alternating series test, but I can not use it because of the $${(-1)^{n+1}}$$ in the denominator.
      Also, I have to find the limit of the series if it does converge.






      sequences-and-series limits convergence divergent-series







      share|cite|improve this question









      $endgroup$




      I have to find out if the series $$sum_{n=1}^infty frac{(-1)^{n+1}}{3n+n(-1)^n}$$ converges. Root test and ratio test did not work out for me. I also tried the alternating series test, but I can not use it because of the $${(-1)^{n+1}}$$ in the denominator.
      Also, I have to find the limit of the series if it does converge.






      sequences-and-series limits convergence divergent-series




      sequences-and-series limits convergence divergent-series






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 8 '18 at 8:54









      Niko BellicNiko Bellic

      82




      82






















          3 Answers
          3






          active

          oldest

          votes


















          1












          $begingroup$

          By observation, the series diverges. In order to prove it diverges, it is sufficient to prove it by proving the grouped series diverges.
          $$sum_{n=1}^infty frac{(-1)^{n+1}}{3n+n(-1)^n}text{ diverges if $sum_{n=1}^infty frac{1}{3(2n-1)-(2n-1)}-frac{1}{3cdot 2n+2n}$diverges.}\
          $$          But we have$$sum_{n=1}^infty frac{1}{3(2n-1)-(2n-1)}-frac{1}{3cdot 2n+2n}\
          =sum_{n=1}^infty frac{2 n+1}{8 n (2 n-1)}=+infty$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank yo for your helpful answer. However, I have trouble understanding if I am allowed to group the terms, because in my opinion this way, I rearrange the summands and there is a theorem that says: If ∑∞n=1an is an absolutely convergent series to the sum s, then any rearrangement of the terms of this series will also converge to the sum s. If ∑∞n=1an is a conditionally convergent series to the sum s, then the terms of this series can be rearranged to converge to any s∈R or to diverge to ±∞. So how can I be sure I didn't make it divergent by rearranging it?
            $endgroup$
            – Niko Bellic
            Dec 8 '18 at 9:37












          • $begingroup$
            @Niko No, grouping is not like rearranging. It is allowed to group terms in convergent series. Try to prove the series diverge By using contraction.
            $endgroup$
            – Kemono Chen
            Dec 8 '18 at 9:41












          • $begingroup$
            Okay, I didn't know that. Thanks. But you say I am allowed to group terms in convergent series. But at the beginning, I don't know whether this one is convergent... So am I allowed anyway to group terms?
            $endgroup$
            – Niko Bellic
            Dec 8 '18 at 9:47












          • $begingroup$
            @Niko Assume it is convergent, the grouped series is convergent. But the grouped one is actually divergent. Hence the original one is not convergent.
            $endgroup$
            – Kemono Chen
            Dec 8 '18 at 9:55












          • $begingroup$
            Sorry, the contraction in above is actually conflict. I'm not so good at English :(
            $endgroup$
            – Kemono Chen
            Dec 8 '18 at 10:00



















          1












          $begingroup$

          It diverges because $$sum_{n=1}^{infty}frac{-1}{nleft(1+3left(-1right)^{n}right)}=sum_{n=1}^{infty}left(frac{1}{2left(2n-1right)}-frac{1}{8n}right)=sum_{n=1}^{infty}frac{2n+1}{8nleft(2n-1right)}$$ obtain $O(frac{1}{n})$ terms.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I had the same idea!
            $endgroup$
            – gimusi
            Dec 8 '18 at 9:10










          • $begingroup$
            @gimusi You did; I hadn't noticed. I made the tracking of powers of $-1$ neater, but that's about it.
            $endgroup$
            – J.G.
            Dec 8 '18 at 9:16










          • $begingroup$
            It seems that I worked by odd and even indices but you worked directly on the expression! That's fine. Bye
            $endgroup$
            – gimusi
            Dec 8 '18 at 9:18



















          -1












          $begingroup$

          HINT



          We have that, separating the sum in odd and even indices



          $$sum_{n=1}^inftyfrac{(-1)^{n+1}}{3n+n(-1)^n}=sum_{k=1}^infty frac{1}{3(2k-1)-2k+1}-sum_{k=1}^infty frac{1}{6k+2k}=$$



          $$=sum_{k=1}^infty frac{1}{4k-2}-sum_{k=1}^infty frac{1}{8k}=sum_{k=1}^infty frac{2k+1}{8k(2k-1)}$$






          share|cite|improve this answer











          $endgroup$













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            3 Answers
            3






            active

            oldest

            votes








            3 Answers
            3






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            1












            $begingroup$

            By observation, the series diverges. In order to prove it diverges, it is sufficient to prove it by proving the grouped series diverges.
            $$sum_{n=1}^infty frac{(-1)^{n+1}}{3n+n(-1)^n}text{ diverges if $sum_{n=1}^infty frac{1}{3(2n-1)-(2n-1)}-frac{1}{3cdot 2n+2n}$diverges.}\
            $$            But we have$$sum_{n=1}^infty frac{1}{3(2n-1)-(2n-1)}-frac{1}{3cdot 2n+2n}\
            =sum_{n=1}^infty frac{2 n+1}{8 n (2 n-1)}=+infty$$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Thank yo for your helpful answer. However, I have trouble understanding if I am allowed to group the terms, because in my opinion this way, I rearrange the summands and there is a theorem that says: If ∑∞n=1an is an absolutely convergent series to the sum s, then any rearrangement of the terms of this series will also converge to the sum s. If ∑∞n=1an is a conditionally convergent series to the sum s, then the terms of this series can be rearranged to converge to any s∈R or to diverge to ±∞. So how can I be sure I didn't make it divergent by rearranging it?
              $endgroup$
              – Niko Bellic
              Dec 8 '18 at 9:37












            • $begingroup$
              @Niko No, grouping is not like rearranging. It is allowed to group terms in convergent series. Try to prove the series diverge By using contraction.
              $endgroup$
              – Kemono Chen
              Dec 8 '18 at 9:41












            • $begingroup$
              Okay, I didn't know that. Thanks. But you say I am allowed to group terms in convergent series. But at the beginning, I don't know whether this one is convergent... So am I allowed anyway to group terms?
              $endgroup$
              – Niko Bellic
              Dec 8 '18 at 9:47












            • $begingroup$
              @Niko Assume it is convergent, the grouped series is convergent. But the grouped one is actually divergent. Hence the original one is not convergent.
              $endgroup$
              – Kemono Chen
              Dec 8 '18 at 9:55












            • $begingroup$
              Sorry, the contraction in above is actually conflict. I'm not so good at English :(
              $endgroup$
              – Kemono Chen
              Dec 8 '18 at 10:00
















            1












            $begingroup$

            By observation, the series diverges. In order to prove it diverges, it is sufficient to prove it by proving the grouped series diverges.
            $$sum_{n=1}^infty frac{(-1)^{n+1}}{3n+n(-1)^n}text{ diverges if $sum_{n=1}^infty frac{1}{3(2n-1)-(2n-1)}-frac{1}{3cdot 2n+2n}$diverges.}\
            $$            But we have$$sum_{n=1}^infty frac{1}{3(2n-1)-(2n-1)}-frac{1}{3cdot 2n+2n}\
            =sum_{n=1}^infty frac{2 n+1}{8 n (2 n-1)}=+infty$$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Thank yo for your helpful answer. However, I have trouble understanding if I am allowed to group the terms, because in my opinion this way, I rearrange the summands and there is a theorem that says: If ∑∞n=1an is an absolutely convergent series to the sum s, then any rearrangement of the terms of this series will also converge to the sum s. If ∑∞n=1an is a conditionally convergent series to the sum s, then the terms of this series can be rearranged to converge to any s∈R or to diverge to ±∞. So how can I be sure I didn't make it divergent by rearranging it?
              $endgroup$
              – Niko Bellic
              Dec 8 '18 at 9:37












            • $begingroup$
              @Niko No, grouping is not like rearranging. It is allowed to group terms in convergent series. Try to prove the series diverge By using contraction.
              $endgroup$
              – Kemono Chen
              Dec 8 '18 at 9:41












            • $begingroup$
              Okay, I didn't know that. Thanks. But you say I am allowed to group terms in convergent series. But at the beginning, I don't know whether this one is convergent... So am I allowed anyway to group terms?
              $endgroup$
              – Niko Bellic
              Dec 8 '18 at 9:47












            • $begingroup$
              @Niko Assume it is convergent, the grouped series is convergent. But the grouped one is actually divergent. Hence the original one is not convergent.
              $endgroup$
              – Kemono Chen
              Dec 8 '18 at 9:55












            • $begingroup$
              Sorry, the contraction in above is actually conflict. I'm not so good at English :(
              $endgroup$
              – Kemono Chen
              Dec 8 '18 at 10:00














            1












            1








            1





            $begingroup$

            By observation, the series diverges. In order to prove it diverges, it is sufficient to prove it by proving the grouped series diverges.
            $$sum_{n=1}^infty frac{(-1)^{n+1}}{3n+n(-1)^n}text{ diverges if $sum_{n=1}^infty frac{1}{3(2n-1)-(2n-1)}-frac{1}{3cdot 2n+2n}$diverges.}\
            $$            But we have$$sum_{n=1}^infty frac{1}{3(2n-1)-(2n-1)}-frac{1}{3cdot 2n+2n}\
            =sum_{n=1}^infty frac{2 n+1}{8 n (2 n-1)}=+infty$$






            share|cite|improve this answer











            $endgroup$



            By observation, the series diverges. In order to prove it diverges, it is sufficient to prove it by proving the grouped series diverges.
            $$sum_{n=1}^infty frac{(-1)^{n+1}}{3n+n(-1)^n}text{ diverges if $sum_{n=1}^infty frac{1}{3(2n-1)-(2n-1)}-frac{1}{3cdot 2n+2n}$diverges.}\
            $$            But we have$$sum_{n=1}^infty frac{1}{3(2n-1)-(2n-1)}-frac{1}{3cdot 2n+2n}\
            =sum_{n=1}^infty frac{2 n+1}{8 n (2 n-1)}=+infty$$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 8 '18 at 9:10

























            answered Dec 8 '18 at 8:59









            Kemono ChenKemono Chen

            3,0721743




            3,0721743












            • $begingroup$
              Thank yo for your helpful answer. However, I have trouble understanding if I am allowed to group the terms, because in my opinion this way, I rearrange the summands and there is a theorem that says: If ∑∞n=1an is an absolutely convergent series to the sum s, then any rearrangement of the terms of this series will also converge to the sum s. If ∑∞n=1an is a conditionally convergent series to the sum s, then the terms of this series can be rearranged to converge to any s∈R or to diverge to ±∞. So how can I be sure I didn't make it divergent by rearranging it?
              $endgroup$
              – Niko Bellic
              Dec 8 '18 at 9:37












            • $begingroup$
              @Niko No, grouping is not like rearranging. It is allowed to group terms in convergent series. Try to prove the series diverge By using contraction.
              $endgroup$
              – Kemono Chen
              Dec 8 '18 at 9:41












            • $begingroup$
              Okay, I didn't know that. Thanks. But you say I am allowed to group terms in convergent series. But at the beginning, I don't know whether this one is convergent... So am I allowed anyway to group terms?
              $endgroup$
              – Niko Bellic
              Dec 8 '18 at 9:47












            • $begingroup$
              @Niko Assume it is convergent, the grouped series is convergent. But the grouped one is actually divergent. Hence the original one is not convergent.
              $endgroup$
              – Kemono Chen
              Dec 8 '18 at 9:55












            • $begingroup$
              Sorry, the contraction in above is actually conflict. I'm not so good at English :(
              $endgroup$
              – Kemono Chen
              Dec 8 '18 at 10:00


















            • $begingroup$
              Thank yo for your helpful answer. However, I have trouble understanding if I am allowed to group the terms, because in my opinion this way, I rearrange the summands and there is a theorem that says: If ∑∞n=1an is an absolutely convergent series to the sum s, then any rearrangement of the terms of this series will also converge to the sum s. If ∑∞n=1an is a conditionally convergent series to the sum s, then the terms of this series can be rearranged to converge to any s∈R or to diverge to ±∞. So how can I be sure I didn't make it divergent by rearranging it?
              $endgroup$
              – Niko Bellic
              Dec 8 '18 at 9:37












            • $begingroup$
              @Niko No, grouping is not like rearranging. It is allowed to group terms in convergent series. Try to prove the series diverge By using contraction.
              $endgroup$
              – Kemono Chen
              Dec 8 '18 at 9:41












            • $begingroup$
              Okay, I didn't know that. Thanks. But you say I am allowed to group terms in convergent series. But at the beginning, I don't know whether this one is convergent... So am I allowed anyway to group terms?
              $endgroup$
              – Niko Bellic
              Dec 8 '18 at 9:47












            • $begingroup$
              @Niko Assume it is convergent, the grouped series is convergent. But the grouped one is actually divergent. Hence the original one is not convergent.
              $endgroup$
              – Kemono Chen
              Dec 8 '18 at 9:55












            • $begingroup$
              Sorry, the contraction in above is actually conflict. I'm not so good at English :(
              $endgroup$
              – Kemono Chen
              Dec 8 '18 at 10:00
















            $begingroup$
            Thank yo for your helpful answer. However, I have trouble understanding if I am allowed to group the terms, because in my opinion this way, I rearrange the summands and there is a theorem that says: If ∑∞n=1an is an absolutely convergent series to the sum s, then any rearrangement of the terms of this series will also converge to the sum s. If ∑∞n=1an is a conditionally convergent series to the sum s, then the terms of this series can be rearranged to converge to any s∈R or to diverge to ±∞. So how can I be sure I didn't make it divergent by rearranging it?
            $endgroup$
            – Niko Bellic
            Dec 8 '18 at 9:37






            $begingroup$
            Thank yo for your helpful answer. However, I have trouble understanding if I am allowed to group the terms, because in my opinion this way, I rearrange the summands and there is a theorem that says: If ∑∞n=1an is an absolutely convergent series to the sum s, then any rearrangement of the terms of this series will also converge to the sum s. If ∑∞n=1an is a conditionally convergent series to the sum s, then the terms of this series can be rearranged to converge to any s∈R or to diverge to ±∞. So how can I be sure I didn't make it divergent by rearranging it?
            $endgroup$
            – Niko Bellic
            Dec 8 '18 at 9:37














            $begingroup$
            @Niko No, grouping is not like rearranging. It is allowed to group terms in convergent series. Try to prove the series diverge By using contraction.
            $endgroup$
            – Kemono Chen
            Dec 8 '18 at 9:41






            $begingroup$
            @Niko No, grouping is not like rearranging. It is allowed to group terms in convergent series. Try to prove the series diverge By using contraction.
            $endgroup$
            – Kemono Chen
            Dec 8 '18 at 9:41














            $begingroup$
            Okay, I didn't know that. Thanks. But you say I am allowed to group terms in convergent series. But at the beginning, I don't know whether this one is convergent... So am I allowed anyway to group terms?
            $endgroup$
            – Niko Bellic
            Dec 8 '18 at 9:47






            $begingroup$
            Okay, I didn't know that. Thanks. But you say I am allowed to group terms in convergent series. But at the beginning, I don't know whether this one is convergent... So am I allowed anyway to group terms?
            $endgroup$
            – Niko Bellic
            Dec 8 '18 at 9:47














            $begingroup$
            @Niko Assume it is convergent, the grouped series is convergent. But the grouped one is actually divergent. Hence the original one is not convergent.
            $endgroup$
            – Kemono Chen
            Dec 8 '18 at 9:55






            $begingroup$
            @Niko Assume it is convergent, the grouped series is convergent. But the grouped one is actually divergent. Hence the original one is not convergent.
            $endgroup$
            – Kemono Chen
            Dec 8 '18 at 9:55














            $begingroup$
            Sorry, the contraction in above is actually conflict. I'm not so good at English :(
            $endgroup$
            – Kemono Chen
            Dec 8 '18 at 10:00




            $begingroup$
            Sorry, the contraction in above is actually conflict. I'm not so good at English :(
            $endgroup$
            – Kemono Chen
            Dec 8 '18 at 10:00











            1












            $begingroup$

            It diverges because $$sum_{n=1}^{infty}frac{-1}{nleft(1+3left(-1right)^{n}right)}=sum_{n=1}^{infty}left(frac{1}{2left(2n-1right)}-frac{1}{8n}right)=sum_{n=1}^{infty}frac{2n+1}{8nleft(2n-1right)}$$ obtain $O(frac{1}{n})$ terms.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              I had the same idea!
              $endgroup$
              – gimusi
              Dec 8 '18 at 9:10










            • $begingroup$
              @gimusi You did; I hadn't noticed. I made the tracking of powers of $-1$ neater, but that's about it.
              $endgroup$
              – J.G.
              Dec 8 '18 at 9:16










            • $begingroup$
              It seems that I worked by odd and even indices but you worked directly on the expression! That's fine. Bye
              $endgroup$
              – gimusi
              Dec 8 '18 at 9:18
















            1












            $begingroup$

            It diverges because $$sum_{n=1}^{infty}frac{-1}{nleft(1+3left(-1right)^{n}right)}=sum_{n=1}^{infty}left(frac{1}{2left(2n-1right)}-frac{1}{8n}right)=sum_{n=1}^{infty}frac{2n+1}{8nleft(2n-1right)}$$ obtain $O(frac{1}{n})$ terms.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              I had the same idea!
              $endgroup$
              – gimusi
              Dec 8 '18 at 9:10










            • $begingroup$
              @gimusi You did; I hadn't noticed. I made the tracking of powers of $-1$ neater, but that's about it.
              $endgroup$
              – J.G.
              Dec 8 '18 at 9:16










            • $begingroup$
              It seems that I worked by odd and even indices but you worked directly on the expression! That's fine. Bye
              $endgroup$
              – gimusi
              Dec 8 '18 at 9:18














            1












            1








            1





            $begingroup$

            It diverges because $$sum_{n=1}^{infty}frac{-1}{nleft(1+3left(-1right)^{n}right)}=sum_{n=1}^{infty}left(frac{1}{2left(2n-1right)}-frac{1}{8n}right)=sum_{n=1}^{infty}frac{2n+1}{8nleft(2n-1right)}$$ obtain $O(frac{1}{n})$ terms.






            share|cite|improve this answer









            $endgroup$



            It diverges because $$sum_{n=1}^{infty}frac{-1}{nleft(1+3left(-1right)^{n}right)}=sum_{n=1}^{infty}left(frac{1}{2left(2n-1right)}-frac{1}{8n}right)=sum_{n=1}^{infty}frac{2n+1}{8nleft(2n-1right)}$$ obtain $O(frac{1}{n})$ terms.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 8 '18 at 9:07









            J.G.J.G.

            26.3k22541




            26.3k22541












            • $begingroup$
              I had the same idea!
              $endgroup$
              – gimusi
              Dec 8 '18 at 9:10










            • $begingroup$
              @gimusi You did; I hadn't noticed. I made the tracking of powers of $-1$ neater, but that's about it.
              $endgroup$
              – J.G.
              Dec 8 '18 at 9:16










            • $begingroup$
              It seems that I worked by odd and even indices but you worked directly on the expression! That's fine. Bye
              $endgroup$
              – gimusi
              Dec 8 '18 at 9:18


















            • $begingroup$
              I had the same idea!
              $endgroup$
              – gimusi
              Dec 8 '18 at 9:10










            • $begingroup$
              @gimusi You did; I hadn't noticed. I made the tracking of powers of $-1$ neater, but that's about it.
              $endgroup$
              – J.G.
              Dec 8 '18 at 9:16










            • $begingroup$
              It seems that I worked by odd and even indices but you worked directly on the expression! That's fine. Bye
              $endgroup$
              – gimusi
              Dec 8 '18 at 9:18
















            $begingroup$
            I had the same idea!
            $endgroup$
            – gimusi
            Dec 8 '18 at 9:10




            $begingroup$
            I had the same idea!
            $endgroup$
            – gimusi
            Dec 8 '18 at 9:10












            $begingroup$
            @gimusi You did; I hadn't noticed. I made the tracking of powers of $-1$ neater, but that's about it.
            $endgroup$
            – J.G.
            Dec 8 '18 at 9:16




            $begingroup$
            @gimusi You did; I hadn't noticed. I made the tracking of powers of $-1$ neater, but that's about it.
            $endgroup$
            – J.G.
            Dec 8 '18 at 9:16












            $begingroup$
            It seems that I worked by odd and even indices but you worked directly on the expression! That's fine. Bye
            $endgroup$
            – gimusi
            Dec 8 '18 at 9:18




            $begingroup$
            It seems that I worked by odd and even indices but you worked directly on the expression! That's fine. Bye
            $endgroup$
            – gimusi
            Dec 8 '18 at 9:18











            -1












            $begingroup$

            HINT



            We have that, separating the sum in odd and even indices



            $$sum_{n=1}^inftyfrac{(-1)^{n+1}}{3n+n(-1)^n}=sum_{k=1}^infty frac{1}{3(2k-1)-2k+1}-sum_{k=1}^infty frac{1}{6k+2k}=$$



            $$=sum_{k=1}^infty frac{1}{4k-2}-sum_{k=1}^infty frac{1}{8k}=sum_{k=1}^infty frac{2k+1}{8k(2k-1)}$$






            share|cite|improve this answer











            $endgroup$


















              -1












              $begingroup$

              HINT



              We have that, separating the sum in odd and even indices



              $$sum_{n=1}^inftyfrac{(-1)^{n+1}}{3n+n(-1)^n}=sum_{k=1}^infty frac{1}{3(2k-1)-2k+1}-sum_{k=1}^infty frac{1}{6k+2k}=$$



              $$=sum_{k=1}^infty frac{1}{4k-2}-sum_{k=1}^infty frac{1}{8k}=sum_{k=1}^infty frac{2k+1}{8k(2k-1)}$$






              share|cite|improve this answer











              $endgroup$
















                -1












                -1








                -1





                $begingroup$

                HINT



                We have that, separating the sum in odd and even indices



                $$sum_{n=1}^inftyfrac{(-1)^{n+1}}{3n+n(-1)^n}=sum_{k=1}^infty frac{1}{3(2k-1)-2k+1}-sum_{k=1}^infty frac{1}{6k+2k}=$$



                $$=sum_{k=1}^infty frac{1}{4k-2}-sum_{k=1}^infty frac{1}{8k}=sum_{k=1}^infty frac{2k+1}{8k(2k-1)}$$






                share|cite|improve this answer











                $endgroup$



                HINT



                We have that, separating the sum in odd and even indices



                $$sum_{n=1}^inftyfrac{(-1)^{n+1}}{3n+n(-1)^n}=sum_{k=1}^infty frac{1}{3(2k-1)-2k+1}-sum_{k=1}^infty frac{1}{6k+2k}=$$



                $$=sum_{k=1}^infty frac{1}{4k-2}-sum_{k=1}^infty frac{1}{8k}=sum_{k=1}^infty frac{2k+1}{8k(2k-1)}$$







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Dec 8 '18 at 9:07

























                answered Dec 8 '18 at 9:02









                gimusigimusi

                92.8k84494




                92.8k84494






























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