0

We have the infinite series: $$ \sum_{n=1}^\infty \frac{1}{n}\frac{1}{2^n} $$ It can be shown that this sum converges by using the ratio test. Thanks to sympy, I know the that it converges to $\ln(2)$. But how can I derive this?, Why is the natural logarithm here.

In general how can we evaluate: $$ \sum_{n=1}^\infty \frac{x^n}{n} $$ For $|x| < 1$

J. W. Tanner
  • 60,406
Siddharth Bhat
  • 21
  • Note that $f(x)=\sum_{n=1}^N \frac{x^n}{n}\implies f'(x)=\sum_{n=1}^{N-1} x^{n-1}$ – user Jan 28 '24 at 21:00
  • $\sum_{n=1}^\infty \frac{x^n}{n}$, $|x| < 1$ is nothing but the Taylor series of $-\ln(1-x)$. If you want to see why (apart from more elementary ways) you can just write $ \sum_{n=1}^\infty \frac{x^n}{n} = \sum_{n=1}^\infty \int_0^x t^{n-1}dt = \int_0^x \sum_{n=1}^\infty t^{n-1}dt = \int_0^x \frac{1}{1-t}dt = -\ln(1-x)$ provided $x \in (0,1)$ and there you can change the order of integration and summation (due to being inside of convergence radius). Similarly for $x \in (-1,0)$ (or just interpret above as integrating over directed intervals). – Presage Jan 28 '24 at 21:01

0 Answers0