# Sum of the series $\sum_{n=0}^{\infty} \frac{1}{4^n(2n+1)}$ [duplicate]

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?

• Try using that $1/(2n+1)=\int_0^1 x^{2n}dx$ and invert sum and integral Mar 12, 2016 at 0:50
• 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)! Mar 12, 2016 at 0:54
• What is interesting is that $\sum_{n=0}^{\infty} \frac{x^n}{2n+1}=\frac{\tanh ^{-1}\left(\sqrt{x}\right)}{\sqrt{x}}$ Mar 12, 2016 at 5:50
• May 28, 2017 at 9:01

Write $$\sum_{n=0}^{\infty} \frac{1}{4^n(2n+1)}=\sum_{n=0}^{\infty} \int_0^1\frac{x^{2n}}{4^n}dx=\int_0^1\frac{dx}{1-x^2/4}=\int_0^1 \bigg(\frac{1}{2(1-x/2)}+ \frac{1}{2(1+x/2)} \bigg)dx=\bigg[-\ln(1-x/2)\bigg]_{x=0}^1+\bigg[\ln(1+x/2)\bigg]_{x=0}^1=\ln3$$
• partial fractions work fine: $\frac{1}{1-x^2/4}=\frac{1}{2(1-x/2)}+\frac{1}{2(1+x/2)}$ Mar 12, 2016 at 1:06