For which values of $a$ does the series $sum_{i=1}^n frac{(-2a)^n}{n^2}$ converge?











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I have this series: $sum_{i=1}^n frac{(-2a)^n}{n^2}$ and the question is for what values of $a$ does the series converge. I'm supposed to write the interval $a$ is in. I first tried the ratio test:



$$ frac{(-2a)^{n+1}}{(n+1)^2} cdot frac{(n)^2}{(-2a)^n} = frac{-2a cdot n^2}{(n+1)^2} = frac{-2a cdot n^2}{n^2+2n+1}.$$



I then divided everything with $n^2$:



$$ frac{-2a cdot n^2}{n^2+2n+1} = frac{-2a}{1+2/n+1/n^2},$$



and when I took the limit as $n to infty$, I ended up with the inequality:



$$ -2a < 1. $$



I divided with $-2$ on both side, and got that $a$ is $-1/2$, but I'm supposed to make an interval, what did I do wrong?






sequences-and-series convergence







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  • $$ -2a < 1 implies a > -frac{1}{2} implies a in left(-frac{1}{2}, inftyright), $$ but this isn't the correct interval, as you should be considering the absolute value of the ratio of $a_{n+1}/a_n$, not the ratio itself.
    – Xander Henderson
    Nov 21 at 15:26















up vote
1
down vote

favorite












I have this series: $sum_{i=1}^n frac{(-2a)^n}{n^2}$ and the question is for what values of $a$ does the series converge. I'm supposed to write the interval $a$ is in. I first tried the ratio test:



$$ frac{(-2a)^{n+1}}{(n+1)^2} cdot frac{(n)^2}{(-2a)^n} = frac{-2a cdot n^2}{(n+1)^2} = frac{-2a cdot n^2}{n^2+2n+1}.$$



I then divided everything with $n^2$:



$$ frac{-2a cdot n^2}{n^2+2n+1} = frac{-2a}{1+2/n+1/n^2},$$



and when I took the limit as $n to infty$, I ended up with the inequality:



$$ -2a < 1. $$



I divided with $-2$ on both side, and got that $a$ is $-1/2$, but I'm supposed to make an interval, what did I do wrong?






sequences-and-series convergence







share|cite|improve this question
























  • $$ -2a < 1 implies a > -frac{1}{2} implies a in left(-frac{1}{2}, inftyright), $$ but this isn't the correct interval, as you should be considering the absolute value of the ratio of $a_{n+1}/a_n$, not the ratio itself.
    – Xander Henderson
    Nov 21 at 15:26













up vote
1
down vote

favorite









up vote
1
down vote

favorite











I have this series: $sum_{i=1}^n frac{(-2a)^n}{n^2}$ and the question is for what values of $a$ does the series converge. I'm supposed to write the interval $a$ is in. I first tried the ratio test:



$$ frac{(-2a)^{n+1}}{(n+1)^2} cdot frac{(n)^2}{(-2a)^n} = frac{-2a cdot n^2}{(n+1)^2} = frac{-2a cdot n^2}{n^2+2n+1}.$$



I then divided everything with $n^2$:



$$ frac{-2a cdot n^2}{n^2+2n+1} = frac{-2a}{1+2/n+1/n^2},$$



and when I took the limit as $n to infty$, I ended up with the inequality:



$$ -2a < 1. $$



I divided with $-2$ on both side, and got that $a$ is $-1/2$, but I'm supposed to make an interval, what did I do wrong?






sequences-and-series convergence







share|cite|improve this question















I have this series: $sum_{i=1}^n frac{(-2a)^n}{n^2}$ and the question is for what values of $a$ does the series converge. I'm supposed to write the interval $a$ is in. I first tried the ratio test:



$$ frac{(-2a)^{n+1}}{(n+1)^2} cdot frac{(n)^2}{(-2a)^n} = frac{-2a cdot n^2}{(n+1)^2} = frac{-2a cdot n^2}{n^2+2n+1}.$$



I then divided everything with $n^2$:



$$ frac{-2a cdot n^2}{n^2+2n+1} = frac{-2a}{1+2/n+1/n^2},$$



and when I took the limit as $n to infty$, I ended up with the inequality:



$$ -2a < 1. $$



I divided with $-2$ on both side, and got that $a$ is $-1/2$, but I'm supposed to make an interval, what did I do wrong?






sequences-and-series convergence




sequences-and-series convergence






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edited Nov 21 at 15:24









Xander Henderson

14.1k103554




14.1k103554










asked Nov 21 at 15:21









Sherya

152




152












  • $$ -2a < 1 implies a > -frac{1}{2} implies a in left(-frac{1}{2}, inftyright), $$ but this isn't the correct interval, as you should be considering the absolute value of the ratio of $a_{n+1}/a_n$, not the ratio itself.
    – Xander Henderson
    Nov 21 at 15:26


















  • $$ -2a < 1 implies a > -frac{1}{2} implies a in left(-frac{1}{2}, inftyright), $$ but this isn't the correct interval, as you should be considering the absolute value of the ratio of $a_{n+1}/a_n$, not the ratio itself.
    – Xander Henderson
    Nov 21 at 15:26
















$$ -2a < 1 implies a > -frac{1}{2} implies a in left(-frac{1}{2}, inftyright), $$ but this isn't the correct interval, as you should be considering the absolute value of the ratio of $a_{n+1}/a_n$, not the ratio itself.
– Xander Henderson
Nov 21 at 15:26




$$ -2a < 1 implies a > -frac{1}{2} implies a in left(-frac{1}{2}, inftyright), $$ but this isn't the correct interval, as you should be considering the absolute value of the ratio of $a_{n+1}/a_n$, not the ratio itself.
– Xander Henderson
Nov 21 at 15:26










2 Answers
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You want to take absolute values in the ratio test. The result should be that it converges if $|-2a| < 1$, i.e.
$-1/2 < a < 1/2$, and diverges if $|-2a| > 1$. But you also need to look at the case $|-2a|=1$, where the ratio test is inconclusive.






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    0
    down vote













    We need that



    $$frac{(-2a)^n}{n^2} to 0 implies |2a|le 1$$



    then




    • for $-1<2a<0$ the series converges by comparison test with $sum frac1{n^2}$


    • for $0<2a<1$ the series converges by alternating series test







    share|cite|improve this answer





















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      2 Answers
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      2 Answers
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      up vote
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      You want to take absolute values in the ratio test. The result should be that it converges if $|-2a| < 1$, i.e.
      $-1/2 < a < 1/2$, and diverges if $|-2a| > 1$. But you also need to look at the case $|-2a|=1$, where the ratio test is inconclusive.






      share|cite|improve this answer

























        up vote
        1
        down vote













        You want to take absolute values in the ratio test. The result should be that it converges if $|-2a| < 1$, i.e.
        $-1/2 < a < 1/2$, and diverges if $|-2a| > 1$. But you also need to look at the case $|-2a|=1$, where the ratio test is inconclusive.






        share|cite|improve this answer























          up vote
          1
          down vote










          up vote
          1
          down vote









          You want to take absolute values in the ratio test. The result should be that it converges if $|-2a| < 1$, i.e.
          $-1/2 < a < 1/2$, and diverges if $|-2a| > 1$. But you also need to look at the case $|-2a|=1$, where the ratio test is inconclusive.






          share|cite|improve this answer












          You want to take absolute values in the ratio test. The result should be that it converges if $|-2a| < 1$, i.e.
          $-1/2 < a < 1/2$, and diverges if $|-2a| > 1$. But you also need to look at the case $|-2a|=1$, where the ratio test is inconclusive.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 21 at 15:27









          Robert Israel

          317k23206457




          317k23206457






















              up vote
              0
              down vote













              We need that



              $$frac{(-2a)^n}{n^2} to 0 implies |2a|le 1$$



              then




              • for $-1<2a<0$ the series converges by comparison test with $sum frac1{n^2}$


              • for $0<2a<1$ the series converges by alternating series test







              share|cite|improve this answer

























                up vote
                0
                down vote













                We need that



                $$frac{(-2a)^n}{n^2} to 0 implies |2a|le 1$$



                then




                • for $-1<2a<0$ the series converges by comparison test with $sum frac1{n^2}$


                • for $0<2a<1$ the series converges by alternating series test







                share|cite|improve this answer























                  up vote
                  0
                  down vote










                  up vote
                  0
                  down vote









                  We need that



                  $$frac{(-2a)^n}{n^2} to 0 implies |2a|le 1$$



                  then




                  • for $-1<2a<0$ the series converges by comparison test with $sum frac1{n^2}$


                  • for $0<2a<1$ the series converges by alternating series test







                  share|cite|improve this answer












                  We need that



                  $$frac{(-2a)^n}{n^2} to 0 implies |2a|le 1$$



                  then




                  • for $-1<2a<0$ the series converges by comparison test with $sum frac1{n^2}$


                  • for $0<2a<1$ the series converges by alternating series test








                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 21 at 15:44









                  gimusi

                  1




                  1






























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