Question

Evaluate:- \[ \lim_{n\to\infty} \left ( \sin \left( {\frac{n}{n^2+1^2}} \right ) +\sin \left ( {\frac{n}{n^2+2^2}} \right ) +\cdots+\sin \left ( {\frac{n}{n^2+n^2}} \right ) \right ) \]
My guess is this converges to \( \pi \over 4 \)

Answer 1

\[\lim_{n\to\infty}\sin\left(\frac{n}{n^2+1}\right)=0\\

\vdots\\\lim_{n\to\infty}\sin\left(\frac{n}{n^2+n^2}\right)=\lim_{n\to\infty}\sin\left(\frac{1}{2n}\right)=0\]
I think it's 0.

Answer 2

we write this as a sum
\[\huge \lim_{n \rightarrow 0} \huge \sum_{k=1}^{n}\sin(\frac{ n }{ n^2+k^2 })\]
\[\huge \lim_{n \rightarrow 0} \huge \sum_{k=1}^{n}\frac{1}{n}\sin \left(\frac{ 1 }{ 1+(\frac{ k }{ n })^2 }\right)\]
evaluating the sum using integrals we have
\[\Delta x=\frac{ b-a }{ n }=\frac{ 1 }{ n }\]
\[x=k/n\]
\[\huge \int\limits_{0}^{1} \sin \left(\ \frac{ 1 }{ 1+x^2 }\right)dx\]

Answer 3

i am still trying to figure out the result

Answer 4

oh i see the problem 1/n shud be in the parenthesis

Answer 5

lets look at the question if there s no sin
we have
\[\int\limits \frac{ 1 }{ 1+x^2 }=\tan^{-1}x\]
so between 0 and 1 its \[\pi/4\]

Answer 6

yeah that's correct ...

Answer 7

you should use the fact that sin(x) ~ x when x->0 and when n->oo, you get that thing.

Answer 8

so we use the advantage of the continuity of the sign function and do the sum inside the sin

Answer 9

yes ... that's what i thought ..