How to find the value of this summation?

Question

How to solve this summation?

$$S=\displaystyle\sum_{n=1}^{\infty}\dfrac{(-1)^n}{n4^{4n+1}}$$

I tried it to convert it into a definite integral but wasn't successful.

Help will be appreciated.

Answer 1

$$\frac{(-1)^n}{n4^{4n+1}}=\frac1{4n}\frac{(-1)^n}{(4^4)^n}=\frac14\frac{\left(-\dfrac1{4^4}\right)^n}n$$

For $-1<x\le1,$ $$\ln(1+x)=x-\frac{x^2}2+\frac{x^3}3-\cdots=-\sum_{r=1}^\infty\frac{(-x)^r}r$$

Can you recognize $x$ here?

See also :

1. Taylor series for $\log(1+x)$ and its convergence

2. Convergence for log 2

Answer 2

Write $$S= \frac14\sum_{n=1}^\infty \frac{(-1)^n}{n(4^4)^n}$$ now notice that $$\sum_{n=0}^\infty (-1)^n x^n = \frac{1}{1+x}$$ which then leads us to $$\sum_{n=0}^\infty\frac{(-1)^n}{(n+1)}x^{n+1} = -\sum_{n=1}^\infty \frac{(-1)^n}{n}x^n = \ln(1+x)$$

Thus $$S=-\frac14 \ln(1+\frac{1}{4^4})$$