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

lim_(n→∞) Σ_(k=1) ^∞ (n/(n^2 + k)) = ?

$$\underset{\boldsymbol{\mathrm{n}}\rightarrow\infty} {\mathrm{lim}}\:\underset{\boldsymbol{\mathrm{k}}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{n}}{\mathrm{n}^{\mathrm{2}} \:+\:\mathrm{k}}\:=\:? \\ $$

Answered by Olaf_Thorendsen last updated on 22/Aug/21

my calculous was false. sorry.

$$\mathrm{my}\:\mathrm{calculous}\:\mathrm{was}\:\mathrm{false}. \\ $$$$\mathrm{sorry}. \\ $$

Commented by puissant last updated on 22/Aug/21

monsieur on ne peut pas utiliser RIEMANN car le ′′k′′ du denominateur n′est pas au carre^� .. en fait c′est ce que je pense..

$${monsieur}\:{on}\:{ne}\:{peut}\:{pas}\:{utiliser} \\ $$$${RIEMANN}\:{car}\:{le}\:''{k}''\:{du}\:{denominateur} \\ $$$${n}'{est}\:{pas}\:{au}\:{carr}\acute {{e}}.. \\ $$$${en}\:{fait}\:{c}'{est}\:{ce}\:{que}\:{je}\:{pense}.. \\ $$

Commented by Olaf_Thorendsen last updated on 22/Aug/21

Exact ! Bien vu !

$$\mathrm{Exact}\:!\:\mathrm{Bien}\:\mathrm{vu}\:! \\ $$

Answered by mindispower last updated on 22/Aug/21

Σ_(k=1) ^n (1/(n(1+(k/n^2 ))))=S 1−x≤(1/(1+x))≤1,∀x≥0 (1/n)−(k/n^3 )≤(1/(n(1+(k/n^2 ))))≤(1/n) Σ_(k=1) ^n ((1/n)−(k/n^3 ))≤Σ_(k=1) ^n (1/(n(1+(k/n^2 ))))≤Σ_(k=1) ^n (1/n) 1−(1/n^3 ).(((n(n+1))/2))≤S≤1⇒S→1

$$\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{n}\left(\mathrm{1}+\frac{{k}}{{n}^{\mathrm{2}} }\right)}={S} \\ $$$$\mathrm{1}−{x}\leqslant\frac{\mathrm{1}}{\mathrm{1}+{x}}\leqslant\mathrm{1},\forall{x}\geqslant\mathrm{0} \\ $$$$\frac{\mathrm{1}}{{n}}−\frac{{k}}{{n}^{\mathrm{3}} }\leqslant\frac{\mathrm{1}}{{n}\left(\mathrm{1}+\frac{{k}}{{n}^{\mathrm{2}} }\right)}\leqslant\frac{\mathrm{1}}{{n}} \\ $$$$\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\frac{\mathrm{1}}{{n}}−\frac{{k}}{{n}^{\mathrm{3}} }\right)\leqslant\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{n}\left(\mathrm{1}+\frac{{k}}{{n}^{\mathrm{2}} }\right)}\leqslant\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{n}} \\ $$$$\mathrm{1}−\frac{\mathrm{1}}{{n}^{\mathrm{3}} }.\left(\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}\right)\leqslant{S}\leqslant\mathrm{1}\Rightarrow{S}\rightarrow\mathrm{1} \\ $$$$ \\ $$$$ \\ $$

Answered by Kamel last updated on 22/Aug/21

Ω=lim_(n→∞) Σ_(k=1) ^n (n/(n^2 + k)) =lim_(n→+∞) n(Ψ(n^2 +n+1)−Ψ(1+n^2 )) =lim_(n→+∞) n(Ln(n^2 +n+1)−Ln(n^2 )) =lim_(t→0^+ ) ((Ln(1+t+t^2 ))/t)=1

$$\Omega=\underset{\boldsymbol{\mathrm{n}}\rightarrow\infty} {\mathrm{lim}}\:\underset{\boldsymbol{\mathrm{k}}=\mathrm{1}} {\overset{{n}} {\sum}}\:\frac{\mathrm{n}}{\mathrm{n}^{\mathrm{2}} \:+\:\mathrm{k}}\: \\ $$$$\:\:\:=\underset{{n}\rightarrow+\infty} {{lim}n}\left(\Psi\left({n}^{\mathrm{2}} +{n}+\mathrm{1}\right)−\Psi\left(\mathrm{1}+{n}^{\mathrm{2}} \right)\right) \\ $$$$\:\:\:=\underset{{n}\rightarrow+\infty} {{lim}n}\left({Ln}\left({n}^{\mathrm{2}} +{n}+\mathrm{1}\right)−{Ln}\left({n}^{\mathrm{2}} \right)\right) \\ $$$$\:\:\:=\underset{{t}\rightarrow\mathrm{0}^{+} } {{lim}}\frac{{Ln}\left(\mathrm{1}+{t}+{t}^{\mathrm{2}} \right)}{{t}}=\mathrm{1} \\ $$

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