A calculus problem by Sandeep Bhardwaj

Denote our sum by $S$ .Let's group the terms: Positives and Negatives:

$\displaystyle S= {\color{#3D99F6}{\frac{1}{1}+\frac{1}{6} + \frac{1}{11} ... }}-( {\color{#D61F06}{\frac{1}{4} + \frac{1}{9} + \frac{1}{14} ...}} )$

Focus on the denominators : they differ by $5$ in each set of terms.So we can rewrite as :

$\displaystyle S=\lim_{k\to \infty} \bigg( {\color{#3D99F6}{\sum_{n=0}^{k} \frac{1}{5n+1}}}- {\color{#D61F06}{\sum_{n=0}^{k} \frac{1}{5n+4}}} \bigg)$

$\displaystyle = \lim_{k\to \infty} \frac{1}{5} \big( \sum_{n=0}^{k} \frac{1}{n+\frac{1}{5}} - \sum_{n=0}^{k} \frac{1}{n+\frac{4}{5}} \big)$

Now Just take out the value at $n=0$ :

$\displaystyle S= \frac{3}{4} + \lim_{k\to \infty} \frac{1}{5} \big( \sum_{n=1}^{k} \frac{1}{n+\frac{1}{5}} - \sum_{n=1}^{k} \frac{1}{n+\frac{4}{5}} \big)$

Recall the definition of the Harmonic Number for any real positive $x$ :

$\displaystyle H_x= \sum_{n=1}^{\infty} \frac{x}{n(x+n)}$

$\displaystyle = \lim_{k\to \infty} \sum_{n=1}^{k} \frac{x}{n(x+n)}$

$\displaystyle = \lim_{k\to \infty} \bigg( \sum_{n=1}^{k} \frac{1}{n} - \sum_{n=1}^{k} \frac{1}{n+x } \bigg)$

Therefore:

$\displaystyle \lim_{k\to \infty} \sum_{n=1}^{k} \frac{1}{n+x } = \lim_{k\to \infty} H_k - H_x$

Substitute that in $S$ :

$\displaystyle S = \frac{3}{4} + \frac{1}{5} \bigg( \lim_{k\to \infty} \sum_{n=1}^{k} \frac{1}{n+\frac{1}{5}} - \lim_{k\to \infty} \sum_{n=1}^{k} \frac{1}{n+\frac{4}{5}} \bigg)$

$\displaystyle = \frac{3}{4} + \frac{1}{5} \bigg( \lim_{k\to \infty} H_k - H_{\frac{1}{5}} - (\lim_{k\to \infty} H_k - H_{\frac{4}{5}} ) \bigg)$

$\displaystyle = \frac{3}{4} + \frac{1}{5} \bigg( H_{\frac{4}{5}} - H_{\frac{1}{5}} \bigg)$

Recall the Reflection formula for the Harmonic Numbers :

$\displaystyle H_{1-z} -H_z = \pi \cot (\pi z ) - \frac{1}{z} + \frac{1}{1-z}$

$\displaystyle => S = \frac{\pi}{5} \cot \frac{\pi}{5}$

$\displaystyle \boxed{= \frac{\pi}{25} \sqrt{25+10\sqrt{5}} }$