Evaluate-lim-n-n-n-2-1-2-n-n-2-2-2-n-n-2-n-2-

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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 ))).

Evaluate

\lim_{n \to \infty} (\frac{n}{n^{2} + 1^{2}} + \frac{n}{n^{2} + 2^{2}} + \dots . . + \frac{n}{n^{2} + n^{2}}) .

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

= r t h t e r m = (\frac{n}{n^{2} + r^{2}})

= l i m n \to \infty \underset{r = o}{\sum^{\infty}} (\frac{\frac{1}{n}}{1 + {(r / n)}^{2}})

= l i \underset{n \to \infty}{m} \underset{r = 0}{\sum^{\infty}} \frac{1}{n} (\frac{1}{1 + {(r / n)}^{2}}

= \int^{1} \underset{0}{} \frac{d \overset{}{x}}{1 + x^{2}}

=∣ {t a n}^{- 1} x \underset{0}{\overset{1}{∣}}

= {t a n}^{- 1} 1 - {t a n}^{- 1} 0

= Π / 4

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

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

h o w y o u w r i t e t h e 3 r d s t e p

(i n t e g r a t i o n o n e) ?

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

its a Rieman sum sir Rahul...

i t s a R i e m a n s u m s i r R a h u l \dots

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.

T h a n k y o u s i r .

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 .

l e t t a k e S_{n} = {[(1 + \frac{1}{n^{2}}) (1 + \frac{2^{2}}{n^{2}}) \dots . (1 + \frac{n^{2}}{n^{2}})]}^{\frac{1}{n}}

l n (S_{n}) = \frac{1}{n} l n (\prod_{k = 1}^{n} (1 + \frac{k^{2}}{n^{2}}))

= \frac{1}{n} \sum_{k = 1}^{n} l n (1 + {(\frac{k}{n})}^{2}) \to \int_{0}^{1} l n (1 + x^{2}) d x

\int_{0}^{1} l n (1 + x^{2}) d x = {[x l n (1 + x^{2})]}_{0}^{1} - \int_{0}^{1} x \frac{2 x}{1 + x^{2}} d x

= l n (2) - 2 \int_{0}^{1} \frac{x^{2} + 1 - 1}{1 + x^{2}} d x

= l n (2) - 2 + 2 \int_{0}^{1} \frac{d x}{1 + x^{2}}

= l n (2) - 2 + + 2 \frac{π}{4} = l n (2) + \frac{π}{2} - 2 s o

{l i m}_{n \to + \infty} S_{n} = l n (2) + \frac{π}{2} - 2 .

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