Help with $ \lim \frac{n\pi}{4} - \left( \frac{n^2}{n^2+1^2} +\frac{n^2}{n^2+2^2} + \cdots + \frac{n^2}{n^2+n^2} \right) $

Question

I have proved that $$\lim_{n\to \infty} \left( \frac{n}{n^2+1^2} +\frac{n}{n^2+2^2} + \cdots + \frac{n}{n^2+n^2} \right) = \frac{\pi}{4},$$ by using the Riemann's sum of $\arctan$ on $[0,1]$. Now I'm interested in computing the next limit $$ L=\lim_{n\to \infty} \left[ \frac{n\pi}{4} - \left( \frac{n^2}{n^2+1^2} +\frac{n^2}{n^2+2^2} + \cdots + \frac{n^2}{n^2+n^2} \right) \right] $$ Note that $$ L= \lim n \left[ \frac{\pi}{4} - \left( \frac{n}{n^2+1^2} +\frac{n}{n^2+2^2} + \cdots + \frac{n}{n^2+n^2} \right) \right] =[\infty \cdot 0].$$ I can't manage to calculate $L$. How can this indeterminate form be solved? Also, by using computer, I suspect that $L$ exists and is equal to $1/4$. Any idea or hint is welcome! (Actually, I dont need a complete solution, just a good start point). Have a nice day you all!

Answer 1

By Taylor's formula $$ \frac{1}{{1 + (x/n)^2 }} = \frac{1}{{1 + (k/n)^2 }} - \frac{1}{n}\frac{{2k/n}}{{(1 + (k/n)^2 )^2 }}(x - k) + \frac{{(3(\xi /n)^2 - 1)}}{{(1 + (\xi /n)^2 )^3 }}\frac{1}{{n^2 }}(x - k)^2 . $$ for $0<x<n$, $0\leq k\leq n$ and some $0<\xi<n$ depending on $x$, $k$ and $n$. Since $$ \left|\frac{{(3(\xi /n)^2 - 1)}}{{(1 + (\xi /n)^2 )^3}} \right|\leq 1 $$ for any $0<\xi<n$, we can write $$ \frac{1}{{1 + (x/n)^2 }} = \frac{1}{{1 + (k/n)^2 }} - \frac{1}{n}\frac{{2k/n}}{{(1 + (k/n)^2 )^2 }}(x - k) + \mathcal{O}\left(\frac{1}{{n^2 }}\right)(x - k)^2 $$ uniformly for $0<x<n$ and $0\leq k\leq n$. Thus $$ \int_{k - 1}^k {\frac{{dx}}{{1 + (x/n)^2 }}} = \frac{1}{{1 + (k/n)^2 }} + \frac{1}{n}\frac{{k/n}}{{(1 + (k/n)^2 )^2 }} + \mathcal{O}\!\left( {\frac{1}{{n^2 }}} \right), $$ and so $$ \frac{1}{n}\sum\limits_{k = 1}^n {\frac{1}{{1 + (k/n)^2 }}} = \frac{1}{n}\int_0^n {\frac{{dx}}{{1 + (x/n)^2 }}} - \frac{1}{{n^2 }}\sum\limits_{k = 1}^n {\frac{{k/n}}{{(1 + (k/n)^2 )^2 }}} + \mathcal{O}\!\left( {\frac{1}{{n^2 }}} \right), $$ i.e., $$ \sum\limits_{k = 1}^n {\frac{{n^2 }}{{n^2 + k^2 }}} = \frac{\pi }{4}n - \frac{1}{n}\sum\limits_{k = 1}^n {\frac{{k/n}}{{(1 + (k/n)^2 )^2 }}} +\mathcal{O}\!\left( {\frac{1}{n}} \right). $$ Finally, $$ \frac{\pi }{4}n - \sum\limits_{k = 1}^n {\frac{{n^2 }}{{n^2 + k^2 }}} = \frac{1}{n}\sum\limits_{k = 1}^n {\frac{{k/n}}{{(1 + (k/n)^2 )^2 }}} + \mathcal{O}\!\left( {\frac{1}{n}} \right) \to \int_0^1 {\frac{x}{{(1 + x^2 )^2 }}dx} = \frac{1}{4}. $$

Answer 2

Let $f(x):=\frac{1}{1+x^2}$ so$$L=\lim_{n\to\infty}L_n,\,L_n:=n\int_0^1f(x)dx-\sum_{i=1}^nf(i/n).$$Taylor-expanding $f$,$$L_n=\sum_{k\ge1}\frac{f^{(k)}(0)}{k!}c_{nk},\,c_{nk}:=n\int_0^1x^kdx-\frac{1}{n^k}\sum_{i=1}^ni^k\sim-\tfrac12$$by Falhaber's formula. So$$L=-\tfrac12\sum_{k\ge1}\frac{f^{(k)}(0)}{k!}=-\tfrac12(f(1)-f(0))=\tfrac14.$$