I am having difficulty with calculating this sum.

By the Ratio Test, the series converges and on the solution key of the last year's exam it written that the sum = $\ln 3$.

I tried known Maclaurin Series but there is a problem every time. For example, when I try the expansion of $\arctan x$ I can't get rid of $(-1)^n.$

Could you help out, or give a hint?

  • $\begingroup$ Try using that $1/(2n+1)=\int_0^1 x^{2n}dx$ and invert sum and integral $\endgroup$
    – Marcel
    Mar 12, 2016 at 0:50
  • $\begingroup$ OK, I will define $f(x) = \frac{x^{2n}}{4^n} \Rightarrow \int f(x)dx = C+ \frac{x^{2n+1}}{4^n(2n+1)}$. I get it now (geometric series)! $\endgroup$
    – frosh
    Mar 12, 2016 at 0:54
  • $\begingroup$ What is interesting is that $\sum_{n=0}^{\infty} \frac{x^n}{2n+1}=\frac{\tanh ^{-1}\left(\sqrt{x}\right)}{\sqrt{x}}$ $\endgroup$ Mar 12, 2016 at 5:50
  • $\begingroup$ See also: Compute $\sum\limits_{n=0}^{\infty}\frac{1}{(2n+1) . 4^n}$. $\endgroup$ May 28, 2017 at 9:01

1 Answer 1


Write $$\sum_{n=0}^{\infty} \frac{1}{4^n(2n+1)}=\sum_{n=0}^{\infty} \int_0^1\frac{x^{2n}}{4^n}dx=\int_0^1\frac{dx}{1-x^2/4}=\int_0^1 \bigg(\frac{1}{2(1-x/2)}+ \frac{1}{2(1+x/2)} \bigg)dx=\bigg[-\ln(1-x/2)\bigg]_{x=0}^1+\bigg[\ln(1+x/2)\bigg]_{x=0}^1=\ln3$$

  • $\begingroup$ partial fractions work fine: $\frac{1}{1-x^2/4}=\frac{1}{2(1-x/2)}+\frac{1}{2(1+x/2)}$ $\endgroup$
    – Marcel
    Mar 12, 2016 at 1:06

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