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Question Number 39443 by rahul 19 last updated on 06/Jul/18

lim_(n→∞) [ (1/(n^2 +1))+ (2/(n^2 +2))+ (3/(n^2 +3))+ ....+(1/(n+1))] = ?

$$\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\:\left[\:\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} +\mathrm{1}}+\:\frac{\mathrm{2}}{\mathrm{n}^{\mathrm{2}} +\mathrm{2}}+\:\frac{\mathrm{3}}{\mathrm{n}^{\mathrm{2}} +\mathrm{3}}+\:....+\frac{\mathrm{1}}{\mathrm{n}+\mathrm{1}}\right]\:=\:? \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 06/Jul/18

very good...this is the way to solve it...

$${very}\:{good}...{this}\:{is}\:{the}\:{way}\:{to}\:{solve}\:{it}... \\ $$

Commented by math khazana by abdo last updated on 06/Jul/18

S_n =Σ_(k=1) ^n (k/(n^2 +k)) we have 1≤k≤n ⇒ 1+n^2 ≤n^2 +k≤n^2 +n ⇒ (1/(n^2 +n)) ≤ (1/(n^2 +k)) ≤ (1/(1+n^2 )) ⇒ Σ_(k=1) ^n (k/(n^2 +n)) ≤ Σ_(k=1) ^n (k/(n^2 +k)) ≤ Σ_(k=1) ^n (k/(n^2 +1)) ⇒ (1/(n^2 +n)) ((n^2 +n)/2) ≤ S_n ≤ (1/(n^2 +1)) ((n^2 +n)/2) ⇒ (1/2) ≤ S_n ≤ ((n^2 +n)/(2(n^2 +1))) ⇒lim_(n→+∞) S_n =(1/2) .

$${S}_{{n}} =\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\frac{{k}}{{n}^{\mathrm{2}} \:+{k}}\:\:{we}\:{have}\:\:\:\:\:\:\:\mathrm{1}\leqslant{k}\leqslant{n}\:\Rightarrow \\ $$$$\mathrm{1}+{n}^{\mathrm{2}} \leqslant{n}^{\mathrm{2}} +{k}\leqslant{n}^{\mathrm{2}} \:+{n}\:\Rightarrow\:\frac{\mathrm{1}}{{n}^{\mathrm{2}} \:+{n}}\:\leqslant\:\frac{\mathrm{1}}{{n}^{\mathrm{2}} \:+{k}}\:\leqslant\:\frac{\mathrm{1}}{\mathrm{1}+{n}^{\mathrm{2}} } \\ $$$$\Rightarrow\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{{k}}{{n}^{\mathrm{2}} +{n}}\:\leqslant\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{{k}}{{n}^{\mathrm{2}} +{k}}\:\leqslant\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{{k}}{{n}^{\mathrm{2}} \:+\mathrm{1}}\:\Rightarrow \\ $$$$\frac{\mathrm{1}}{{n}^{\mathrm{2}} \:+{n}}\:\frac{{n}^{\mathrm{2}} \:+{n}}{\mathrm{2}}\:\leqslant\:{S}_{{n}} \:\leqslant\:\frac{\mathrm{1}}{{n}^{\mathrm{2}} +\mathrm{1}}\:\frac{{n}^{\mathrm{2}} \:+{n}}{\mathrm{2}}\:\Rightarrow \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\:\leqslant\:{S}_{{n}} \:\leqslant\:\:\frac{{n}^{\mathrm{2}} \:+{n}}{\mathrm{2}\left({n}^{\mathrm{2}} +\mathrm{1}\right)}\:\Rightarrow{lim}_{{n}\rightarrow+\infty} \:{S}_{{n}} =\frac{\mathrm{1}}{\mathrm{2}}\:. \\ $$

Commented by abdo mathsup 649 cc last updated on 07/Jul/18

thanks.

$${thanks}. \\ $$

Answered by ajfour last updated on 06/Jul/18

Let n^2 =N L=lim_(N→∞) NΣ_(r=1) ^(√N) (((r/N)((1/N)))/(1+(r/N))) =lim_(N→∞) N∫_0 ^( 1/(√N)) ((xdx)/(1+x)) =lim_(N→∞) N[(1/(√N))−ln (1+(1/(√N)))] =lim_(N→∞) N[(1/(√N))−(1/(√N))+(1/(2N))−....] ⇒ L = (1/2) .

$${Let}\:{n}^{\mathrm{2}} ={N} \\ $$$${L}=\underset{{N}\rightarrow\infty} {\mathrm{lim}}{N}\underset{{r}=\mathrm{1}} {\overset{\sqrt{{N}}} {\sum}}\:\frac{\left({r}/{N}\right)\left(\frac{\mathrm{1}}{{N}}\right)}{\mathrm{1}+\frac{{r}}{{N}}} \\ $$$$\:\:\:=\underset{{N}\rightarrow\infty} {\mathrm{lim}}{N}\int_{\mathrm{0}} ^{\:\:\mathrm{1}/\sqrt{{N}}} \:\frac{{xdx}}{\mathrm{1}+{x}} \\ $$$$\:\:=\underset{{N}\rightarrow\infty} {\mathrm{lim}}{N}\left[\frac{\mathrm{1}}{\sqrt{{N}}}−\mathrm{ln}\:\left(\mathrm{1}+\frac{\mathrm{1}}{\sqrt{{N}}}\right)\right] \\ $$$$\:\:=\underset{{N}\rightarrow\infty} {\mathrm{lim}}\:{N}\left[\frac{\mathrm{1}}{\sqrt{{N}}}−\frac{\mathrm{1}}{\sqrt{{N}}}+\frac{\mathrm{1}}{\mathrm{2}{N}}−....\right] \\ $$$$\Rightarrow\:\:\:\:\:{L}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:. \\ $$

Commented by rahul 19 last updated on 06/Jul/18

Ans. given is 1/2

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