Question and Answers Forum

All Questions      Topic List

Limits Questions

Previous in All Question      Next in All Question      

Previous in Limits      Next in Limits      

Question Number 34367 by rahul 19 last updated on 05/May/18

Evaluate   lim_(n→∞) ((n/(n^2 +1^2 ))+(n/(n^2 +2^2 ))+.....+(n/(n^2 +n^2 ))).

$$\boldsymbol{\mathrm{Evaluate}}\: \\ $$$$\mathrm{lim}_{\mathrm{n}\rightarrow\infty} \left(\frac{{n}}{{n}^{\mathrm{2}} +\mathrm{1}^{\mathrm{2}} }+\frac{{n}}{{n}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} }+.....+\frac{{n}}{{n}^{\mathrm{2}} +{n}^{\mathrm{2}} }\right). \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 05/May/18

=rth term =((n/(n^2 +r^2 )))  =lim n→∞Σ_(r=o) ^∞ (((1/n)/(1+(r/n)^2  )))  =lim_(n→∞)  Σ_(r=0) ^∞ (1/n)((1/(1+(r/n)^2 ))  =∫^1  _0 (dx^ /(1+x^2 ))  =∣tan^(−1) x∣_0 ^1   =tan^(−1) 1−tan^(−1) 0  =Π/4

$$={rth}\:{term}\:=\left(\frac{{n}}{{n}^{\mathrm{2}} +{r}^{\mathrm{2}} }\right) \\ $$$$={lim}\:{n}\rightarrow\infty\underset{{r}={o}} {\overset{\infty} {\sum}}\left(\frac{\frac{\mathrm{1}}{{n}}}{\mathrm{1}+\left({r}/{n}\right)^{\mathrm{2}} \:}\right) \\ $$$$={li}\underset{{n}\rightarrow\infty} {{m}}\:\underset{{r}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}}\left(\frac{\mathrm{1}}{\mathrm{1}+\left({r}/{n}\right)^{\mathrm{2}} }\right. \\ $$$$=\overset{\mathrm{1}} {\int}\underset{\mathrm{0}} {\:}\frac{{d}\overset{} {{x}}}{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$$=\mid{tan}^{−\mathrm{1}} {x}\underset{\mathrm{0}} {\overset{\mathrm{1}} {\mid}} \\ $$$$={tan}^{−\mathrm{1}} \mathrm{1}−{tan}^{−\mathrm{1}} \mathrm{0} \\ $$$$=\Pi/\mathrm{4} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Commented by rahul 19 last updated on 05/May/18

how you write the 3rd step  ( integration one ) ?

$${how}\:{you}\:{write}\:{the}\:\mathrm{3}{rd}\:{step} \\ $$$$\left(\:{integration}\:{one}\:\right)\:? \\ $$

Commented by math khazana by abdo last updated on 05/May/18

its a Rieman sum sir Rahul...

$${its}\:{a}\:{Rieman}\:{sum}\:{sir}\:{Rahul}... \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 05/May/18

Commented by tanmay.chaudhury50@gmail.com last updated on 05/May/18

Commented by tanmay.chaudhury50@gmail.com last updated on 05/May/18

Commented by tanmay.chaudhury50@gmail.com last updated on 05/May/18

Commented by tanmay.chaudhury50@gmail.com last updated on 05/May/18

Commented by rahul 19 last updated on 05/May/18

Thank you sir.

$$\mathscr{T}{hank}\:{you}\:{sir}. \\ $$

Commented by abdo mathsup 649 cc last updated on 05/May/18

let take  S_n =[ (1+(1/n^2 ))(1+(2^2 /n^2 ))....(1+(n^2 /n^2 ))]^(1/n)   ln(S_n ) =(1/n)ln( Π_(k=1) ^n (1+(k^2 /n^2 )))  =(1/n) Σ_(k=1) ^n  ln(1+((k/n))^2 )→∫_0 ^1 ln(1+x^2 )dx  ∫_0 ^1   ln(1+x^2 )dx = [x ln(1+x^2 )]_0 ^1  −∫_0 ^1  x ((2x)/(1+x^2 ))dx  =ln(2)  −2 ∫_0 ^1   ((x^2 +1−1)/(1+x^2 ))dx  =ln(2) −2  +2 ∫_0 ^1     (dx/(1+x^2 ))  =ln(2) −2 + +2 (π/4) = ln(2) +(π/2) −2 so  lim_(n→+∞)   S_n  = ln(2)+ (π/2) −2 .

$${let}\:{take}\:\:{S}_{{n}} =\left[\:\left(\mathrm{1}+\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\right)\left(\mathrm{1}+\frac{\mathrm{2}^{\mathrm{2}} }{{n}^{\mathrm{2}} }\right)....\left(\mathrm{1}+\frac{{n}^{\mathrm{2}} }{{n}^{\mathrm{2}} }\right)\right]^{\frac{\mathrm{1}}{{n}}} \\ $$$${ln}\left({S}_{{n}} \right)\:=\frac{\mathrm{1}}{{n}}{ln}\left(\:\prod_{{k}=\mathrm{1}} ^{{n}} \left(\mathrm{1}+\frac{{k}^{\mathrm{2}} }{{n}^{\mathrm{2}} }\right)\right) \\ $$$$=\frac{\mathrm{1}}{{n}}\:\sum_{{k}=\mathrm{1}} ^{{n}} \:{ln}\left(\mathrm{1}+\left(\frac{{k}}{{n}}\right)^{\mathrm{2}} \right)\rightarrow\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right){dx} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\:{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right){dx}\:=\:\left[{x}\:{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\right]_{\mathrm{0}} ^{\mathrm{1}} \:−\int_{\mathrm{0}} ^{\mathrm{1}} \:{x}\:\frac{\mathrm{2}{x}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\ $$$$={ln}\left(\mathrm{2}\right)\:\:−\mathrm{2}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{x}^{\mathrm{2}} +\mathrm{1}−\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\ $$$$={ln}\left(\mathrm{2}\right)\:−\mathrm{2}\:\:+\mathrm{2}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$$={ln}\left(\mathrm{2}\right)\:−\mathrm{2}\:+\:+\mathrm{2}\:\frac{\pi}{\mathrm{4}}\:=\:{ln}\left(\mathrm{2}\right)\:+\frac{\pi}{\mathrm{2}}\:−\mathrm{2}\:{so} \\ $$$${lim}_{{n}\rightarrow+\infty} \:\:{S}_{{n}} \:=\:{ln}\left(\mathrm{2}\right)+\:\frac{\pi}{\mathrm{2}}\:−\mathrm{2}\:. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com