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Question Number 174630 by infinityaction last updated on 06/Aug/22

lim_(n→∞) Σ_(r=1) ^∞ (r/(n^2 +r))

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

Answered by mnjuly1970 last updated on 06/Aug/22

lim_(n→∞) {(1/(1+n^( 2) )) +(2/(2 +n^( 2) )) +(3/(3+n^( 2) )) +...(n/(n+n^( 2) )) =a_( n) } ≤ (1/(1+n^( 2) )) +(2/(1+n^( 2) )) +...+(n/(1+n^( 2) )) = ((n(n+1))/(2(1+n^( 2) ))) (1) a_n ≥(1/(n+n^( 2) )) +(2/(n+n^( 2) )) +...+(n/(n+n^( 2) )) = ((n(n+1))/(2n(1+n))) (1/2) ≤ a_( n) ≤ ((n(n+1))/(2(1+n^2 ))) lim_(n→) (a_( n) )= (1/2)

$$\:\:\:\:\:{lim}_{{n}\rightarrow\infty} \left\{\frac{\mathrm{1}}{\mathrm{1}+{n}^{\:\mathrm{2}} }\:+\frac{\mathrm{2}}{\mathrm{2}\:+{n}^{\:\mathrm{2}} }\:+\frac{\mathrm{3}}{\mathrm{3}+{n}^{\:\mathrm{2}} }\:+...\frac{{n}}{{n}+{n}^{\:\mathrm{2}} }\:={a}_{\:{n}} \right\} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\leqslant\:\frac{\mathrm{1}}{\mathrm{1}+{n}^{\:\mathrm{2}} }\:+\frac{\mathrm{2}}{\mathrm{1}+{n}^{\:\mathrm{2}} }\:+...+\frac{{n}}{\mathrm{1}+{n}^{\:\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:=\:\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}\left(\mathrm{1}+{n}^{\:\mathrm{2}} \right)}\:\:\:\:\left(\mathrm{1}\right) \\ $$$$\:\:\:\:\:\:\:\:{a}_{{n}} \:\geqslant\frac{\mathrm{1}}{{n}+{n}^{\:\mathrm{2}} }\:+\frac{\mathrm{2}}{{n}+{n}^{\:\mathrm{2}} }\:+...+\frac{{n}}{{n}+{n}^{\:\mathrm{2}} } \\ $$$$=\:\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}{n}\left(\mathrm{1}+{n}\right)}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}}\:\leqslant\:{a}_{\:{n}} \:\leqslant\:\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}\left(\mathrm{1}+{n}\:^{\mathrm{2}} \right)} \\ $$$$\:\:\:\:\:\:{lim}_{{n}\rightarrow} \:\left({a}_{\:{n}} \right)=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$

Commented by infinityaction last updated on 06/Aug/22

thanks sir

$${thanks}\:{sir} \\ $$

Commented by mnjuly1970 last updated on 06/Aug/22

you are welcome sir...

$$\:{you}\:{are}\:{welcome}\:{sir}... \\ $$

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