Evaluating the limit $\displaystyle \lim_{n\to\infty} \left[ \frac{n}{n^2+1}+ \frac{n}{n^2+2^2} + \ldots + \frac{n}{n^2+n^2} \right]$

I have a question about the following solution:

We may write it in the form:

$$ \frac{1}{n} \left[ \frac{1}{1+(\frac{1}{n})^2} + \ldots + \frac{1}{1+(\frac{n}{n})^2} \right] $$

Somehow I need to figure out that the limit is actually the Riemann sum of $\frac{1}{1 + x^2}$ on $[0,1]$ for $\pi = 0 < \frac{1}{n} < \ldots < \frac{n}{n}$.

Can you explain to me to reach this conclusion?


The number you seek is \begin{equation} L = \lim_{n\to\infty}\sum_{k=1}^n\frac{n}{n^2+k^2} = \lim_{n\to\infty}\sum_{k=1}^n\frac{1}{1+\left(\frac{k}{n}\right)^2}\cdot\frac{1}{n}\tag{1} \end{equation} Now, letting \begin{align*} \Delta x &= \frac{1}{n} & x_k &= \frac{k}{n} \end{align*} allows us to rewrite (1) as $$ L = \lim_{n\to\infty}\sum_{k=1}^n\frac{1}{1+x_k^2}\Delta x $$ If you're familiar with the standard construction of the Riemann integral, you will recognize this as $$ L=\int_0^1\frac{1}{1+x^2}\,dx $$

  • $\begingroup$ @AlonAlon Yes of course. Fixed. $\endgroup$ – Brian Fitzpatrick Jan 15 '15 at 19:22
  • $\begingroup$ Thank you Brian (and all)! $\endgroup$ – AlonAlon Jan 15 '15 at 19:30

Well you already gave the answer can't figure out what's the problem? You just say: Let $f(x)=\dfrac{1}{1+x^2}$.

$$\lim\limits_{n\to +\infty}\sum\limits_{k=1}^n\frac{n^2}{n^2+k^2}=\lim\limits_{n\to +\infty}\frac{1-0}{n}\sum\limits_{k=1}^nf\left( 0+\frac{1-0}{n}k\right)=\int\limits_0^1f(t)\mathrm dt$$


$$\frac1n\sum_{k=1}^n\frac1{1+\left(\frac kn\right)^2}\xrightarrow[n\to\infty]{}\int_0^1\frac{dx}{1+x^2}\left(=\frac\pi4\;.\;\;Why?\right)$$

Just note that the above sum is

$$\sum_{k=1}^n f\left(\frac kn\right)(x_{k+1}-x_k)\,,\,\,\text{with}\;\;f(x)=\frac1{1+x^2}\;$$

and the partition of the unit interval

$$\left\{0<\frac1n<\frac2n\ldots<\frac nn=1\right\}\;,\;\;\text{and}\;\;c_k:=\frac kn$$


Hints : First look at the coefficient $\frac{1}{n}$ in front of the expression inside brackets. The $1$ at the top of the fraction is $1-0$, where $1$ and $0$ are the end points of the interval $[0,1]$. The $n$ at the bottom would be the length of a subinterval of $[0,1]$ when you partition $[0,1]$ into $n$ equal parts.

Then look at each individual term inside brackets. For example, $$ \frac{1}{1+(\frac{3}{n})^2} $$ would be the value of the function $\frac{1}{1+x^2}$ when $x$ is $3$ times the length of a subinterval of length $\frac{1}{n}$ added from the starting point $0$.

Can you now figure out that this is the Riemann sum you had to recognize ? What the expression really represents is a sum of areas of rectangles of base $\frac{1}{n}$ and height $\frac{1}{1+(\frac{i}{n})^2}$. Here you approach the area under the curve $\frac{1}{1+x^2}$ over $[0,1]$ by taking the right end points of the subintervals of an equally distributed partition.


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