L-lim-n-1-n-2-1-2-2-n-2-2-2-n-n-2-n-2-

Question Number 144396 by SOMEDAVONG last updated on 25/Jun/21

L = \lim_{n \to + \propto} (\frac{1}{n^{2} + 1^{2}} + \frac{2}{n^{2} + 2^{2}} + . . + \frac{n}{n^{2} + n^{2}}) = ?

Commented by Canebulok last updated on 25/Jun/21

S o l u t i o n :

\Rightarrow \lim_{n \to + \infty} (\underset{k = 1}{\sum^{n}} \frac{k}{n^{2} + k^{2}}) = L

\Rightarrow \lim_{n \to + \infty} (\frac{1}{n}) (\underset{k = 1}{\sum^{n}} \frac{n k}{n^{2} + k^{2}}) = L

\Rightarrow \lim_{n \to + \infty} (\frac{1}{n}) (\underset{k = 1}{\sum^{n}} \frac{1}{\frac{n}{k} + \frac{k}{n}}) = L

∴

\Rightarrow \int_{0}^{1} \frac{1}{(\frac{1}{x} + x)} d x = L

\Rightarrow \int_{0}^{1} \frac{x}{x^{2} + 1} d x = [\frac{L n (∣ x^{2} + 1 ∣)}{2} ∣_{0}^{1}]

\Rightarrow L = \frac{L n (∣ 2 ∣)}{2}

Answered by mathmax by abdo last updated on 25/Jun/21

L = \lim_{n \to + \infty} \sum_{k = 1}^{n} \frac{k}{n^{2} + k^{2}} = \lim_{n \to + \infty} \frac{\frac{k}{n}}{n + \frac{k^{2}}{n}}

= \lim_{n \to + \infty} \frac{1}{n} \times \frac{\frac{k}{n}}{1 + {(\frac{k}{n})}^{2}} = \int_{0}^{1} \frac{x}{1 + x^{2}} dx = \frac{1}{2} {[\log (1 + x^{2})]}_{0}^{1}

= \frac{1}{2} \log 2