Jun 7, 2021 - 03:34

An infinite series problem.

By Vamshi Jandhyala in mathematics

June 12, 2021

Problem

Solution

We have

$$ \begin{aligned} \sum_{n=1}^\infty \frac{1}{n(2n+1)} &= 2\left(\frac{1}{2} - \frac{1}{3} + \frac{1}{4} - \frac{1}{5} + \cdots \right) \\
&= 2 \int_0^1 x - x^2 + x^3 + \dots dx \\
&= 2 \int_0^1 \frac{x}{1+x} dx \\
&= 2 \int_0^1 1 - \frac{1}{1+x} dx \\
&= 2 - \ln{2} \end{aligned} $$

We have

$$ \begin{aligned} \sum_{n=1}^\infty \frac{1}{n(3n+1)} &= 3\left(\frac{1}{3} - \frac{1}{4} + \frac{1}{6} - \frac{1}{7} + \cdots \right) \\
&= 3 \int_0^1 x^2 + x^5 + x^8 + \dots - (x^3 + x^6 + x^9 + \dots) dx \\
&= 3 \int_0^1 \frac{x^2}{1 - x^3} - \frac{x^3}{1 - x^3}dx \\
&= 3 - \frac{\pi}{2\sqrt{3}} - \frac{3\ln{3}}{2} \end{aligned} $$

We have

$$ \begin{aligned} \sum_{n=1}^\infty \frac{1}{n(4n+1)} &= 4\left(\frac{1}{4} - \frac{1}{5} + \frac{1}{8} - \frac{1}{9} + \cdots \right) \\
&= 4 \int_0^1 x^3 + x^7 + \dots - (x^4 + x^8 + \dots) dx \\
&= 4 \int_0^1 \frac{x^3}{1 - x^4} - \frac{x^4}{1 - x^4}dx \\
&= 4 - 3 \ln{2} - \frac{\pi}{2} \end{aligned} $$

The following identities can also be proved using a similar approach illustrated previously

We have

$$ \begin{aligned} \frac{1}{1} - \frac{1}{4} + \frac{1}{7} - \cdots &= \int_0^1 1 - x^3 + x^6 - \cdots dx \\
&= \int_0^1 \frac{1}{1+x^3} dx \\
&= \frac{1}{3}\left( \frac{\pi}{\sqrt{3}} + \ln{2} \right) \end{aligned} $$

$$ \begin{aligned} \frac{1}{2} - \frac{1}{5} + \frac{1}{8} - \cdots &= \int_0^1 x - x^4 + x^7 - \cdots dx \\
&= \int_0^1 \frac{x}{1+x^3} dx \\
&= \frac{1}{3}\left( \frac{\pi}{\sqrt{3}} - \ln{2} \right) \end{aligned} $$