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

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

• In this forum, we ask that YOU give us your attempts and thoughts first. Commented Nov 27, 2015 at 13:49
• 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). Commented Nov 27, 2015 at 13:52
• @PeterWoolfitt: Indeed if you multiply by $\frac{1}{2}$, this is $\tanh^{-1} \frac{1}{2}$. Please feel free to post an Answer. Commented Nov 27, 2015 at 13:56
• @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). Commented Nov 27, 2015 at 13:58

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 $\ln 3$.