Stuck finding the sum of two series












1












$begingroup$


This is the first time I ever make a post on Stack Exchange (and the last time I stop lurking it!), so apologies in advance if I caused any problems.



I am completely stuck at finding the sum of the two following series:



$$sum_{n=1}^∞(2n+1)x^n$$
$$sum_{n=0}^∞frac{x^n}{(n+1)2^n}$$



I understand that I should be using this series to find out their value: $$sum_{n=0}^∞x^n=frac{1}{1-x}$$



I already used it for a couple of simpler series, but I cannot get my head around the two ones above






calculus sequences-and-series







share|cite|improve this question









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  • $begingroup$
    Be careful with your lower limits! What you need here (in conjunction with @DonAntonio's answer) is $sum_{n=1}^infty x^n$, not $sum_{n=0}^infty x^n$.
    $endgroup$
    – TonyK
    Dec 23 '18 at 21:37
















1












$begingroup$


This is the first time I ever make a post on Stack Exchange (and the last time I stop lurking it!), so apologies in advance if I caused any problems.



I am completely stuck at finding the sum of the two following series:



$$sum_{n=1}^∞(2n+1)x^n$$
$$sum_{n=0}^∞frac{x^n}{(n+1)2^n}$$



I understand that I should be using this series to find out their value: $$sum_{n=0}^∞x^n=frac{1}{1-x}$$



I already used it for a couple of simpler series, but I cannot get my head around the two ones above






calculus sequences-and-series







share|cite|improve this question









$endgroup$












  • $begingroup$
    Be careful with your lower limits! What you need here (in conjunction with @DonAntonio's answer) is $sum_{n=1}^infty x^n$, not $sum_{n=0}^infty x^n$.
    $endgroup$
    – TonyK
    Dec 23 '18 at 21:37














1












1








1





$begingroup$


This is the first time I ever make a post on Stack Exchange (and the last time I stop lurking it!), so apologies in advance if I caused any problems.



I am completely stuck at finding the sum of the two following series:



$$sum_{n=1}^∞(2n+1)x^n$$
$$sum_{n=0}^∞frac{x^n}{(n+1)2^n}$$



I understand that I should be using this series to find out their value: $$sum_{n=0}^∞x^n=frac{1}{1-x}$$



I already used it for a couple of simpler series, but I cannot get my head around the two ones above






calculus sequences-and-series







share|cite|improve this question









$endgroup$




This is the first time I ever make a post on Stack Exchange (and the last time I stop lurking it!), so apologies in advance if I caused any problems.



I am completely stuck at finding the sum of the two following series:



$$sum_{n=1}^∞(2n+1)x^n$$
$$sum_{n=0}^∞frac{x^n}{(n+1)2^n}$$



I understand that I should be using this series to find out their value: $$sum_{n=0}^∞x^n=frac{1}{1-x}$$



I already used it for a couple of simpler series, but I cannot get my head around the two ones above






calculus sequences-and-series




calculus sequences-and-series






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 23 '18 at 18:54









LightsongLightsong

318




318












  • $begingroup$
    Be careful with your lower limits! What you need here (in conjunction with @DonAntonio's answer) is $sum_{n=1}^infty x^n$, not $sum_{n=0}^infty x^n$.
    $endgroup$
    – TonyK
    Dec 23 '18 at 21:37


















  • $begingroup$
    Be careful with your lower limits! What you need here (in conjunction with @DonAntonio's answer) is $sum_{n=1}^infty x^n$, not $sum_{n=0}^infty x^n$.
    $endgroup$
    – TonyK
    Dec 23 '18 at 21:37
















$begingroup$
Be careful with your lower limits! What you need here (in conjunction with @DonAntonio's answer) is $sum_{n=1}^infty x^n$, not $sum_{n=0}^infty x^n$.
$endgroup$
– TonyK
Dec 23 '18 at 21:37




$begingroup$
Be careful with your lower limits! What you need here (in conjunction with @DonAntonio's answer) is $sum_{n=1}^infty x^n$, not $sum_{n=0}^infty x^n$.
$endgroup$
– TonyK
Dec 23 '18 at 21:37










2 Answers
2






active

oldest

votes


















0












$begingroup$


  1. Set $f(x) = sum_{n=1}^infty (2n+1)x^n$. Then
    $$
    f(t^2)
    = sum_{n=1}^infty (2n+1)t^{2n}
    = sum_{n=1}^infty frac{d}{dt}t^{2n+1}
    = frac{d}{dt} sum_{n=1}^infty t^{2n+1}
    = frac{d}{dt} left( t sum_{n=1}^infty (t^2)^n right)
    = frac{d}{dt} left( frac{t^3}{1-t^2} right) \
    = frac{3t^2(1-t^2)-t^3(-2t)}{(1-t^2)^2}
    = frac{3t^2-t^4}{(1-t^2)^2}.
    $$
    Thus,
    $$
    f(x) = frac{3x-x^2}{(1-x)^2}.
    $$


  2. Set $g(x) = sum_{n=0}^infty frac{x^n}{(n+1)2^n}.$ Then,
    $$
    frac{d}{dx} left( x , g(x) right)
    = frac{d}{dx} left( 2 sum_{n=0}^infty frac{(x/2)^{n+1}}{(n+1)} right)
    = 2 sum_{n=0}^infty frac{1}{2} (x/2)^n
    = frac{1}{1-x/2}
    = frac{2}{2-x}.
    $$
    Therefore, for some constant $C$,
    $$x , g(x) = C - 2 ln(2-x).$$
    The constant can be determined by $0 = 0 , g(0) = C - 2 ln 2,$ i.e. $C = 2 ln 2.$
    Thus,
    $$g(x) = frac{2 ln 2}{x} - 2 frac{ln(2-x)}{x} = 2 frac{ln 2 - ln(2-x)}{x} = 2 frac{ln(2/(2-x))}{x}.$$







share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you! Now I got it.
    $endgroup$
    – Lightsong
    Dec 24 '18 at 11:06



















0












$begingroup$

For example:



$$frac1{1-x}=sum_{n=0}^infty x^nimplies frac1{(1-x)^2}=sum_{n=1}^infty nx^{n-1}impliesfrac x{(1-x)^2}=sum_{n=1}^infty nx^n;,;;|x|<1$$



and now do some calculations with $;(2n+1)x^n=2nx^n+x^n;$, with the help of arithmetic of series (limits)



For the other one integrate:



$$frac1{1-x}=sum_{n=0}^infty x^nimplies -log(1-x)=sum_{n=0}^inftyfrac{x^{n+1}}{n+1};,;;|x|<1$$



You may also want to consider $;frac x2;$ in the last part...






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks! Though I still don't quite understand it: For the 1st, what do you mean by the "arithmetic of series(limits)"? We haven't seen limits in power series so far, only for calculating the radius of convergence. For the 2nd one, then I would need to play with 1/(x*2^n) * −log(1−x) = $$sum_{n=0}^∞frac{x^n}{(n+1)2^n}$$? I don't know how to advance from there.
    $endgroup$
    – Lightsong
    Dec 23 '18 at 20:03










  • $begingroup$
    But I think you may have seen that sum of convergent series is convergent, so $;sum(a_n+b_n)=sum a_n+sum b_n;$ when each series separately converges. That what I meant. For the second one check the series with $;frac x2;$ instead, meaning: $$-log(1-2x)=sumfrac{left(frac x2right)^{n+1}}{{n+1}};ldots$$
    $endgroup$
    – DonAntonio
    Dec 23 '18 at 20:25












  • $begingroup$
    You might have mentioned that what you are doing in the first line is differentiating...
    $endgroup$
    – TonyK
    Dec 23 '18 at 21:32










  • $begingroup$
    @TonyK Perhaps, but if the OP can't figure out that by himself then maybe this stuff is not for him. I also say in the next line "For the other one integrate" ...I suppose that's a huge hint, too.
    $endgroup$
    – DonAntonio
    Dec 23 '18 at 21:42












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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$


  1. Set $f(x) = sum_{n=1}^infty (2n+1)x^n$. Then
    $$
    f(t^2)
    = sum_{n=1}^infty (2n+1)t^{2n}
    = sum_{n=1}^infty frac{d}{dt}t^{2n+1}
    = frac{d}{dt} sum_{n=1}^infty t^{2n+1}
    = frac{d}{dt} left( t sum_{n=1}^infty (t^2)^n right)
    = frac{d}{dt} left( frac{t^3}{1-t^2} right) \
    = frac{3t^2(1-t^2)-t^3(-2t)}{(1-t^2)^2}
    = frac{3t^2-t^4}{(1-t^2)^2}.
    $$
    Thus,
    $$
    f(x) = frac{3x-x^2}{(1-x)^2}.
    $$


  2. Set $g(x) = sum_{n=0}^infty frac{x^n}{(n+1)2^n}.$ Then,
    $$
    frac{d}{dx} left( x , g(x) right)
    = frac{d}{dx} left( 2 sum_{n=0}^infty frac{(x/2)^{n+1}}{(n+1)} right)
    = 2 sum_{n=0}^infty frac{1}{2} (x/2)^n
    = frac{1}{1-x/2}
    = frac{2}{2-x}.
    $$
    Therefore, for some constant $C$,
    $$x , g(x) = C - 2 ln(2-x).$$
    The constant can be determined by $0 = 0 , g(0) = C - 2 ln 2,$ i.e. $C = 2 ln 2.$
    Thus,
    $$g(x) = frac{2 ln 2}{x} - 2 frac{ln(2-x)}{x} = 2 frac{ln 2 - ln(2-x)}{x} = 2 frac{ln(2/(2-x))}{x}.$$







share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you! Now I got it.
    $endgroup$
    – Lightsong
    Dec 24 '18 at 11:06
















0












$begingroup$


  1. Set $f(x) = sum_{n=1}^infty (2n+1)x^n$. Then
    $$
    f(t^2)
    = sum_{n=1}^infty (2n+1)t^{2n}
    = sum_{n=1}^infty frac{d}{dt}t^{2n+1}
    = frac{d}{dt} sum_{n=1}^infty t^{2n+1}
    = frac{d}{dt} left( t sum_{n=1}^infty (t^2)^n right)
    = frac{d}{dt} left( frac{t^3}{1-t^2} right) \
    = frac{3t^2(1-t^2)-t^3(-2t)}{(1-t^2)^2}
    = frac{3t^2-t^4}{(1-t^2)^2}.
    $$
    Thus,
    $$
    f(x) = frac{3x-x^2}{(1-x)^2}.
    $$


  2. Set $g(x) = sum_{n=0}^infty frac{x^n}{(n+1)2^n}.$ Then,
    $$
    frac{d}{dx} left( x , g(x) right)
    = frac{d}{dx} left( 2 sum_{n=0}^infty frac{(x/2)^{n+1}}{(n+1)} right)
    = 2 sum_{n=0}^infty frac{1}{2} (x/2)^n
    = frac{1}{1-x/2}
    = frac{2}{2-x}.
    $$
    Therefore, for some constant $C$,
    $$x , g(x) = C - 2 ln(2-x).$$
    The constant can be determined by $0 = 0 , g(0) = C - 2 ln 2,$ i.e. $C = 2 ln 2.$
    Thus,
    $$g(x) = frac{2 ln 2}{x} - 2 frac{ln(2-x)}{x} = 2 frac{ln 2 - ln(2-x)}{x} = 2 frac{ln(2/(2-x))}{x}.$$







share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you! Now I got it.
    $endgroup$
    – Lightsong
    Dec 24 '18 at 11:06














0












0








0





$begingroup$


  1. Set $f(x) = sum_{n=1}^infty (2n+1)x^n$. Then
    $$
    f(t^2)
    = sum_{n=1}^infty (2n+1)t^{2n}
    = sum_{n=1}^infty frac{d}{dt}t^{2n+1}
    = frac{d}{dt} sum_{n=1}^infty t^{2n+1}
    = frac{d}{dt} left( t sum_{n=1}^infty (t^2)^n right)
    = frac{d}{dt} left( frac{t^3}{1-t^2} right) \
    = frac{3t^2(1-t^2)-t^3(-2t)}{(1-t^2)^2}
    = frac{3t^2-t^4}{(1-t^2)^2}.
    $$
    Thus,
    $$
    f(x) = frac{3x-x^2}{(1-x)^2}.
    $$


  2. Set $g(x) = sum_{n=0}^infty frac{x^n}{(n+1)2^n}.$ Then,
    $$
    frac{d}{dx} left( x , g(x) right)
    = frac{d}{dx} left( 2 sum_{n=0}^infty frac{(x/2)^{n+1}}{(n+1)} right)
    = 2 sum_{n=0}^infty frac{1}{2} (x/2)^n
    = frac{1}{1-x/2}
    = frac{2}{2-x}.
    $$
    Therefore, for some constant $C$,
    $$x , g(x) = C - 2 ln(2-x).$$
    The constant can be determined by $0 = 0 , g(0) = C - 2 ln 2,$ i.e. $C = 2 ln 2.$
    Thus,
    $$g(x) = frac{2 ln 2}{x} - 2 frac{ln(2-x)}{x} = 2 frac{ln 2 - ln(2-x)}{x} = 2 frac{ln(2/(2-x))}{x}.$$







share|cite|improve this answer









$endgroup$




  1. Set $f(x) = sum_{n=1}^infty (2n+1)x^n$. Then
    $$
    f(t^2)
    = sum_{n=1}^infty (2n+1)t^{2n}
    = sum_{n=1}^infty frac{d}{dt}t^{2n+1}
    = frac{d}{dt} sum_{n=1}^infty t^{2n+1}
    = frac{d}{dt} left( t sum_{n=1}^infty (t^2)^n right)
    = frac{d}{dt} left( frac{t^3}{1-t^2} right) \
    = frac{3t^2(1-t^2)-t^3(-2t)}{(1-t^2)^2}
    = frac{3t^2-t^4}{(1-t^2)^2}.
    $$
    Thus,
    $$
    f(x) = frac{3x-x^2}{(1-x)^2}.
    $$


  2. Set $g(x) = sum_{n=0}^infty frac{x^n}{(n+1)2^n}.$ Then,
    $$
    frac{d}{dx} left( x , g(x) right)
    = frac{d}{dx} left( 2 sum_{n=0}^infty frac{(x/2)^{n+1}}{(n+1)} right)
    = 2 sum_{n=0}^infty frac{1}{2} (x/2)^n
    = frac{1}{1-x/2}
    = frac{2}{2-x}.
    $$
    Therefore, for some constant $C$,
    $$x , g(x) = C - 2 ln(2-x).$$
    The constant can be determined by $0 = 0 , g(0) = C - 2 ln 2,$ i.e. $C = 2 ln 2.$
    Thus,
    $$g(x) = frac{2 ln 2}{x} - 2 frac{ln(2-x)}{x} = 2 frac{ln 2 - ln(2-x)}{x} = 2 frac{ln(2/(2-x))}{x}.$$








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share|cite|improve this answer










answered Dec 23 '18 at 21:25









md2perpemd2perpe

8,73411129




8,73411129












  • $begingroup$
    Thank you! Now I got it.
    $endgroup$
    – Lightsong
    Dec 24 '18 at 11:06


















  • $begingroup$
    Thank you! Now I got it.
    $endgroup$
    – Lightsong
    Dec 24 '18 at 11:06
















$begingroup$
Thank you! Now I got it.
$endgroup$
– Lightsong
Dec 24 '18 at 11:06




$begingroup$
Thank you! Now I got it.
$endgroup$
– Lightsong
Dec 24 '18 at 11:06











0












$begingroup$

For example:



$$frac1{1-x}=sum_{n=0}^infty x^nimplies frac1{(1-x)^2}=sum_{n=1}^infty nx^{n-1}impliesfrac x{(1-x)^2}=sum_{n=1}^infty nx^n;,;;|x|<1$$



and now do some calculations with $;(2n+1)x^n=2nx^n+x^n;$, with the help of arithmetic of series (limits)



For the other one integrate:



$$frac1{1-x}=sum_{n=0}^infty x^nimplies -log(1-x)=sum_{n=0}^inftyfrac{x^{n+1}}{n+1};,;;|x|<1$$



You may also want to consider $;frac x2;$ in the last part...






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks! Though I still don't quite understand it: For the 1st, what do you mean by the "arithmetic of series(limits)"? We haven't seen limits in power series so far, only for calculating the radius of convergence. For the 2nd one, then I would need to play with 1/(x*2^n) * −log(1−x) = $$sum_{n=0}^∞frac{x^n}{(n+1)2^n}$$? I don't know how to advance from there.
    $endgroup$
    – Lightsong
    Dec 23 '18 at 20:03










  • $begingroup$
    But I think you may have seen that sum of convergent series is convergent, so $;sum(a_n+b_n)=sum a_n+sum b_n;$ when each series separately converges. That what I meant. For the second one check the series with $;frac x2;$ instead, meaning: $$-log(1-2x)=sumfrac{left(frac x2right)^{n+1}}{{n+1}};ldots$$
    $endgroup$
    – DonAntonio
    Dec 23 '18 at 20:25












  • $begingroup$
    You might have mentioned that what you are doing in the first line is differentiating...
    $endgroup$
    – TonyK
    Dec 23 '18 at 21:32










  • $begingroup$
    @TonyK Perhaps, but if the OP can't figure out that by himself then maybe this stuff is not for him. I also say in the next line "For the other one integrate" ...I suppose that's a huge hint, too.
    $endgroup$
    – DonAntonio
    Dec 23 '18 at 21:42
















0












$begingroup$

For example:



$$frac1{1-x}=sum_{n=0}^infty x^nimplies frac1{(1-x)^2}=sum_{n=1}^infty nx^{n-1}impliesfrac x{(1-x)^2}=sum_{n=1}^infty nx^n;,;;|x|<1$$



and now do some calculations with $;(2n+1)x^n=2nx^n+x^n;$, with the help of arithmetic of series (limits)



For the other one integrate:



$$frac1{1-x}=sum_{n=0}^infty x^nimplies -log(1-x)=sum_{n=0}^inftyfrac{x^{n+1}}{n+1};,;;|x|<1$$



You may also want to consider $;frac x2;$ in the last part...






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks! Though I still don't quite understand it: For the 1st, what do you mean by the "arithmetic of series(limits)"? We haven't seen limits in power series so far, only for calculating the radius of convergence. For the 2nd one, then I would need to play with 1/(x*2^n) * −log(1−x) = $$sum_{n=0}^∞frac{x^n}{(n+1)2^n}$$? I don't know how to advance from there.
    $endgroup$
    – Lightsong
    Dec 23 '18 at 20:03










  • $begingroup$
    But I think you may have seen that sum of convergent series is convergent, so $;sum(a_n+b_n)=sum a_n+sum b_n;$ when each series separately converges. That what I meant. For the second one check the series with $;frac x2;$ instead, meaning: $$-log(1-2x)=sumfrac{left(frac x2right)^{n+1}}{{n+1}};ldots$$
    $endgroup$
    – DonAntonio
    Dec 23 '18 at 20:25












  • $begingroup$
    You might have mentioned that what you are doing in the first line is differentiating...
    $endgroup$
    – TonyK
    Dec 23 '18 at 21:32










  • $begingroup$
    @TonyK Perhaps, but if the OP can't figure out that by himself then maybe this stuff is not for him. I also say in the next line "For the other one integrate" ...I suppose that's a huge hint, too.
    $endgroup$
    – DonAntonio
    Dec 23 '18 at 21:42














0












0








0





$begingroup$

For example:



$$frac1{1-x}=sum_{n=0}^infty x^nimplies frac1{(1-x)^2}=sum_{n=1}^infty nx^{n-1}impliesfrac x{(1-x)^2}=sum_{n=1}^infty nx^n;,;;|x|<1$$



and now do some calculations with $;(2n+1)x^n=2nx^n+x^n;$, with the help of arithmetic of series (limits)



For the other one integrate:



$$frac1{1-x}=sum_{n=0}^infty x^nimplies -log(1-x)=sum_{n=0}^inftyfrac{x^{n+1}}{n+1};,;;|x|<1$$



You may also want to consider $;frac x2;$ in the last part...






share|cite|improve this answer









$endgroup$



For example:



$$frac1{1-x}=sum_{n=0}^infty x^nimplies frac1{(1-x)^2}=sum_{n=1}^infty nx^{n-1}impliesfrac x{(1-x)^2}=sum_{n=1}^infty nx^n;,;;|x|<1$$



and now do some calculations with $;(2n+1)x^n=2nx^n+x^n;$, with the help of arithmetic of series (limits)



For the other one integrate:



$$frac1{1-x}=sum_{n=0}^infty x^nimplies -log(1-x)=sum_{n=0}^inftyfrac{x^{n+1}}{n+1};,;;|x|<1$$



You may also want to consider $;frac x2;$ in the last part...







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 23 '18 at 18:59









DonAntonioDonAntonio

180k1495233




180k1495233












  • $begingroup$
    Thanks! Though I still don't quite understand it: For the 1st, what do you mean by the "arithmetic of series(limits)"? We haven't seen limits in power series so far, only for calculating the radius of convergence. For the 2nd one, then I would need to play with 1/(x*2^n) * −log(1−x) = $$sum_{n=0}^∞frac{x^n}{(n+1)2^n}$$? I don't know how to advance from there.
    $endgroup$
    – Lightsong
    Dec 23 '18 at 20:03










  • $begingroup$
    But I think you may have seen that sum of convergent series is convergent, so $;sum(a_n+b_n)=sum a_n+sum b_n;$ when each series separately converges. That what I meant. For the second one check the series with $;frac x2;$ instead, meaning: $$-log(1-2x)=sumfrac{left(frac x2right)^{n+1}}{{n+1}};ldots$$
    $endgroup$
    – DonAntonio
    Dec 23 '18 at 20:25












  • $begingroup$
    You might have mentioned that what you are doing in the first line is differentiating...
    $endgroup$
    – TonyK
    Dec 23 '18 at 21:32










  • $begingroup$
    @TonyK Perhaps, but if the OP can't figure out that by himself then maybe this stuff is not for him. I also say in the next line "For the other one integrate" ...I suppose that's a huge hint, too.
    $endgroup$
    – DonAntonio
    Dec 23 '18 at 21:42


















  • $begingroup$
    Thanks! Though I still don't quite understand it: For the 1st, what do you mean by the "arithmetic of series(limits)"? We haven't seen limits in power series so far, only for calculating the radius of convergence. For the 2nd one, then I would need to play with 1/(x*2^n) * −log(1−x) = $$sum_{n=0}^∞frac{x^n}{(n+1)2^n}$$? I don't know how to advance from there.
    $endgroup$
    – Lightsong
    Dec 23 '18 at 20:03










  • $begingroup$
    But I think you may have seen that sum of convergent series is convergent, so $;sum(a_n+b_n)=sum a_n+sum b_n;$ when each series separately converges. That what I meant. For the second one check the series with $;frac x2;$ instead, meaning: $$-log(1-2x)=sumfrac{left(frac x2right)^{n+1}}{{n+1}};ldots$$
    $endgroup$
    – DonAntonio
    Dec 23 '18 at 20:25












  • $begingroup$
    You might have mentioned that what you are doing in the first line is differentiating...
    $endgroup$
    – TonyK
    Dec 23 '18 at 21:32










  • $begingroup$
    @TonyK Perhaps, but if the OP can't figure out that by himself then maybe this stuff is not for him. I also say in the next line "For the other one integrate" ...I suppose that's a huge hint, too.
    $endgroup$
    – DonAntonio
    Dec 23 '18 at 21:42
















$begingroup$
Thanks! Though I still don't quite understand it: For the 1st, what do you mean by the "arithmetic of series(limits)"? We haven't seen limits in power series so far, only for calculating the radius of convergence. For the 2nd one, then I would need to play with 1/(x*2^n) * −log(1−x) = $$sum_{n=0}^∞frac{x^n}{(n+1)2^n}$$? I don't know how to advance from there.
$endgroup$
– Lightsong
Dec 23 '18 at 20:03




$begingroup$
Thanks! Though I still don't quite understand it: For the 1st, what do you mean by the "arithmetic of series(limits)"? We haven't seen limits in power series so far, only for calculating the radius of convergence. For the 2nd one, then I would need to play with 1/(x*2^n) * −log(1−x) = $$sum_{n=0}^∞frac{x^n}{(n+1)2^n}$$? I don't know how to advance from there.
$endgroup$
– Lightsong
Dec 23 '18 at 20:03












$begingroup$
But I think you may have seen that sum of convergent series is convergent, so $;sum(a_n+b_n)=sum a_n+sum b_n;$ when each series separately converges. That what I meant. For the second one check the series with $;frac x2;$ instead, meaning: $$-log(1-2x)=sumfrac{left(frac x2right)^{n+1}}{{n+1}};ldots$$
$endgroup$
– DonAntonio
Dec 23 '18 at 20:25






$begingroup$
But I think you may have seen that sum of convergent series is convergent, so $;sum(a_n+b_n)=sum a_n+sum b_n;$ when each series separately converges. That what I meant. For the second one check the series with $;frac x2;$ instead, meaning: $$-log(1-2x)=sumfrac{left(frac x2right)^{n+1}}{{n+1}};ldots$$
$endgroup$
– DonAntonio
Dec 23 '18 at 20:25














$begingroup$
You might have mentioned that what you are doing in the first line is differentiating...
$endgroup$
– TonyK
Dec 23 '18 at 21:32




$begingroup$
You might have mentioned that what you are doing in the first line is differentiating...
$endgroup$
– TonyK
Dec 23 '18 at 21:32












$begingroup$
@TonyK Perhaps, but if the OP can't figure out that by himself then maybe this stuff is not for him. I also say in the next line "For the other one integrate" ...I suppose that's a huge hint, too.
$endgroup$
– DonAntonio
Dec 23 '18 at 21:42




$begingroup$
@TonyK Perhaps, but if the OP can't figure out that by himself then maybe this stuff is not for him. I also say in the next line "For the other one integrate" ...I suppose that's a huge hint, too.
$endgroup$
– DonAntonio
Dec 23 '18 at 21:42


















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