Find the sum of the series : $$1+\frac{1}{3}\cdot\frac{1}{4}+\frac{1}{5}\cdot\frac{1}{4^2}+\frac{1}{7}\cdot\frac{1}{4^3}+\cdots$$

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    $\begingroup$ In this forum, we ask that YOU give us your attempts and thoughts first. $\endgroup$
    – GEdgar
    Commented Nov 27, 2015 at 13:49
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    $\begingroup$ huh, apparently this is very similar to the Taylor series for arctanh, but I don't know how to see that without knowing (though it also has another super nice form I won't spoil - so maybe there's another way of doing it). $\endgroup$ Commented Nov 27, 2015 at 13:52
  • $\begingroup$ @PeterWoolfitt: Indeed if you multiply by $\frac{1}{2}$, this is $\tanh^{-1} \frac{1}{2}$. Please feel free to post an Answer. $\endgroup$
    – hardmath
    Commented Nov 27, 2015 at 13:56
  • $\begingroup$ @hardmath well, all I did was use WolframAlpha - I don't like posting answers like that when it seems other somehow more legitimate answers will appear. I didn't mean to monopolize this avenue of answer - if you or anyone else wants to post an answer using the arctanh idea, please feel free (The current answer by Gyumin Roh is pretty great). $\endgroup$ Commented Nov 27, 2015 at 13:58

1 Answer 1


This can be transformed to $$\sum_{n=1}^{\infty} \frac{2}{(2n-1)2^{2n-1}}$$

Let $$f(x)=\sum_{n=1}^{\infty} \frac{x^{2n-1}}{(2n-1)}$$

Then, we have $f'(x)=\sum_{n=1}^{\infty} x^{2n-2} = \frac{1}{1-x^2}$.

Therefore, we have $$f(x)=\int \frac{1}{1-x^2} = \frac{1}{2} \ln \frac{x+1}{1-x}+C$$ It is clear that $C=0$.

Now plugging $x=\frac{1}{2}$ in this equation, we have $\sum_{n=1}^{\infty} \frac{1}{(2n-1)2^{2n-1}} = \frac{1}{2} \ln 3$, so the desired answer is double that number, or $\ln 3$.