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Question Number 57385 by Tinkutara last updated on 03/Apr/19

Commented by Tinkutara last updated on 03/Apr/19

How to prove d?

Answered by einsteindrmaths@hotmail.fr last updated on 03/Apr/19

$$S_n<\underset{0}{\sum ^{\mathrm{\infty }}}\frac{n}{n^2+kn+k^2}={lim}_{n\to \mathrm{\infty }}\underset{0}{\sum ^n}\frac{n}{n^2+kn+k^2}={lim}_{n\to \mathrm{\infty }}\frac{1}{n}\underset{0}{\sum ^n}\frac{1}{1+\frac{k}{n}+{\left(\frac{k}{n}\right)}^2}$$$$=\underset{0}{\stackrel{1}{\int }}\frac{1}{1+x+x^2}=\underset{0}{\stackrel{1}{\int }}\frac{1}{{(x+\frac{1}{2})}^2+\frac{3}{4}}=\frac{2}{\sqrt{3}}{tan}^{-1}[\frac{2}{\sqrt{3}}+\frac{2}{2\sqrt{3}}]-\frac{2}{\sqrt{3}}{tan}^{-1}\left[\frac{2}{2\sqrt{3}}\right]<\frac{2}{\sqrt{3}}{tan}^{-1}(\sqrt{3)}$$

Commented by Tinkutara last updated on 03/Apr/19

Sir please prove d option, I already got a

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