Evalute-the-sum-k-1-n-k-pi-n-2-arctan-kpi-n-2-

Question Number 164196 by HongKing last updated on 15/Jan/22

Evalute the sum :

\underset{k = 1}{\sum^{n}} k {(\frac{π}{n})}^{2} \arctan {(\frac{k π}{n})}^{2}

Answered by mathmax by abdo last updated on 15/Jan/22

S_{n} = \frac{π}{n} \sum_{k = 1}^{n} (\frac{k π}{n}) \arctan {(\frac{k π}{n})}^{2} \Rightarrow \lim_{n \to + \infty} S_{n} = \int_{0}^{π} xarctan (x^{2}) dx

=_{by parts} {[\frac{x^{2}}{2} \arctan (x^{2})]}_{0}^{π} - \int_{0}^{π} \frac{x^{2}}{2} \frac{2 x}{1 + x^{4}} dx

= \frac{π^{2}}{2} \arctan (π^{2}) - \int_{0}^{π} \frac{x^{3}}{1 + x^{4}} dx and

\int_{0}^{π} \frac{x^{3}}{1 + x^{4}} dx = {[\frac{1}{4} \ln (1 + x^{4})]}_{0}^{π} = \frac{1}{4} \ln (1 + π^{4}) \Rightarrow

\lim_{n \to + \infty} S_{n} = \frac{π^{2}}{2} \arctan (π^{2}) - \frac{1}{4} \ln (1 + π^{4})

Commented by HongKing last updated on 15/Jan/22

thank you so much my dear Sir