lim-n-k-1-n-n-2-k-

Question Number 151767 by mathdanisur last updated on 22/Aug/21

\lim_{n \to \infty} \underset{k = 1}{\sum^{\infty}} \frac{n}{n^{2} + k} = ?

Answered by Olaf_Thorendsen last updated on 22/Aug/21

my calculous was false .

sorry .

Commented by puissant last updated on 22/Aug/21

m o n s i e u r o n n e p e u t p a s u t i l i s e r

R I E M A N N c a r l e^{″} k^{″} d u d e n o m i n a t e u r

n^{'} e s t p a s a u c a r r \overset{´}{e} . .

e n f a i t c^{'} e s t c e q u e j e p e n s e . .

Commented by Olaf_Thorendsen last updated on 22/Aug/21

Exact! Bien vu!

Answered by mindispower last updated on 22/Aug/21

\underset{k = 1}{\sum^{n}} \frac{1}{n (1 + \frac{k}{n^{2}})} = S

1 - x ⩽ \frac{1}{1 + x} ⩽ 1, \forall x ⩾ 0

\frac{1}{n} - \frac{k}{n^{3}} ⩽ \frac{1}{n (1 + \frac{k}{n^{2}})} ⩽ \frac{1}{n}

\underset{k = 1}{\sum^{n}} (\frac{1}{n} - \frac{k}{n^{3}}) ⩽ \underset{k = 1}{\sum^{n}} \frac{1}{n (1 + \frac{k}{n^{2}})} ⩽ \underset{k = 1}{\sum^{n}} \frac{1}{n}

1 - \frac{1}{n^{3}} . (\frac{n (n + 1)}{2}) ⩽ S ⩽ 1 \Rightarrow S \to 1

Answered by Kamel last updated on 22/Aug/21

Ω = \lim_{n \to \infty} \underset{k = 1}{\sum^{n}} \frac{n}{n^{2} + k}

= \underset{n \to + \infty}{l i m n} (Ψ (n^{2} + n + 1) - Ψ (1 + n^{2}))

= \underset{n \to + \infty}{l i m n} (L n (n^{2} + n + 1) - L n (n^{2}))

= \underset{t \to 0^{+}}{l i m} \frac{L n (1 + t + t^{2})}{t} = 1

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