How to find sum of the infinite series $\sum_{n=1}^{\infty} \frac{1}{ n(2n+1)}$

Question

$$\frac{1}{1 \times3} + \frac{1}{2\times5}+\frac{1}{3\times7} + \frac{1}{4\times9}+\cdots $$

How to find sum of this series?

I tried this: its $n$th term will be = $\frac{1}{n}-\frac{2}{2n+1}$; after that I am not able to solve this.

Answer 1

Your series can be written as $$ \frac{H_{x}}{2x} = \sum_{k=1}^\infty \frac{1}{k\cdot2(k+x)}\, , $$ with $x=\frac12$. $H_{\frac12}$ is given here: Harmonic numbers. $$ H_{\frac{1}{2}} = 2 -2\ln{2}, $$ which is your result...

Answer 2

First, we can rewrite the partial sum as an integral $$\sum_{n=1}^N \frac{1}{n(2n+1)} = 2\sum_{n=1}^N \left(\frac{1}{2n} - \frac{1}{2n+1}\right) = 2\sum_{n=1}^N \int_0^1 (z^{2n-1} - z^{2n}) dz\\ = 2 \int_0^1 z(1-z)\left(\sum_{n=0}^{N-1} z^{2n}\right) dz = 2 \int_0^1 \frac{z}{1+z}( 1 - z^{2N} ) dz $$ Notice the $N$ dependence piece on RHS can be bounded from above $$\left| 2 \int_0^1 \frac{z}{1+z} z^{2N} dz \right| < 2 \int_0^1 z^{2N} dz = \frac{2}{2N+1} \to 0 \quad\text{ as }\quad N \to \infty $$ We have $$\sum_{n=1}^\infty \frac{1}{n(2n+1)} =\lim_{N\to\infty}\sum_{n=1}^N \frac{1}{n(2n+1)} = 2 \int_0^1 \frac{z}{1+z} dz = 2 (1 - \log 2)$$

Answer 3

Let $f(x)=\sum_{n=1}^{\infty} \frac{x^{2n+1}}{ n(2n+1)}$. Then we have $$f'(x)=\sum_{n=1}^{\infty} \frac{x^{2n}}{ n}=-\log(1-x^2).$$ Hence since f(0)=0, the sum is equal to \begin{align} s&=-\int_0^1\log(1-x^2)dx\\ &=-2\int_0^{\pi/2}\log(\cos x) \cos x dx\\ &=-2I \end{align} To solve this integral, $I$, note first that $\int_0^{\pi/2}\log(\sin x) \cos x dx=-1$. Thus \begin{align} I&=\int_0^{\pi/2}\log(\cos x) \cos x dx-\int_0^{\pi/2}\log(\sin x) \cos x dx+\int_0^{\pi/2}\log(\sin x) \cos x dx\\ &=\int_0^{\pi/2}\log(\cot x) \cos x dx-1\\ &=\int_0^{\pi/2}\Big(\log(\cot x) \cos x+\sec x -\sec x \Big)dx-1\\ &=\lim_{a\to \pi/2}\int_0^{a}\Big(\log(\cot x) \cos x-\sec x +\sec x \Big)dx-1\\ &=\lim_{a\to \pi/2}\Big(\log(\cot x) \sin x +\log [\cos \frac x2 + \sin \frac x2]-\log [\cos \frac x2 -\sin \frac x2] \Big)-1\\ &=\log2-1 \end{align}

... I should still add more ...