Considering the sum as a Riemann sum, evaluate $\lim_{n\to\infty}\sum_{k=1}^{n}\frac{k}{n^2+k^2} .$
Evaluate $\lim_{n\to\infty}\sum_{k=1}^{n}\frac{k}{n^2+k^2}$
-
2$\dfrac{1}{n}\dfrac{\frac{k}{n}}{1+\left(\frac{k}{n}\right)^2}$ – 2012-12-12
3 Answers
$\sum_{k=1}^n\frac{k}{n^2+k^2}=\frac{1}{n^2}\sum_{k=1}^n\frac{k}{1+\left(\frac{k}{n}\right)^2}=\frac{1}{n}\sum_{k=1}^n\frac{\frac{k}{n}}{1+\left(\frac{k}{n}\right)^2}\xrightarrow [n\to\infty]{}\int_0^1\frac{x}{1+x^2}\,dx=\ldots$
A more general approach. When you have sums of the form $\sum_{k=0}^nf(k,n)$ and want to test convergence you may want to use the definition of Riemann Integrals. Choose your favorite partition, mine is \begin{equation}\mathcal{P}=\left\{ 0=x_0
Another approach, more quantitative than recognizing the limit as a Riemann sum, is to estimate using integral approximations.
First note that $f(x)=\frac{x}{n^2+x^2}$ has derivative $f'(x)=\frac{n^2-x^2}{(n^2+x^2)^2}$ so $f$ is increasing when $x\in[0,n]$. Thus, we can estimate $\sum f(k)$ by the same kind of approach that is used in the integral test, except we deal with an increasing $f$ rather than decreasing.
For $1\leq k\leq n-1$, we have $\int_{k-1}^kf(x)\,dx\leq\int_{k-1}^kf(k)\,dx=f(k)=\int_k^{k+1}f(k)\,dx\leq\int_k^{k+1}f(x)\,dx$
Summing from $k=2$ to $k=n-1$ gives
$\int_1^{n-1}f(x)\,dx\leq\sum_{k=2}^{n-1}f(k)\leq\int_2^nf(x)\,dx$
Adding this to the boundary cases
$f(1)\leq f(1)\leq\int_1^2f(x)\,dx$
and
$\int_{n-1}^nf(x)\,dx\leq f(n)\leq f(n)$ gives
$f(1)+\int_1^{n}f(x)\,dx\leq\sum_{k=1}^{n}f(k)\leq f(n)+\int_1^nf(x)\,dx$
But
$\int_1^nf(x)\,dx=\frac{1}{2}\log\frac{2n^2}{n^2+1}$
so we have the estimates
$\frac{1}{n^2+1}+\frac{1}{2}\log\frac{2n^2}{n^2+1}\leq S_n\leq\frac{1}{2n}+\frac{1}{2}\log\frac{2n^2}{n^2+1}$
Letting $n\to\infty$ gives $S_n\to\frac{1}{2}\log 2$.
-
1@Santiago: Thanks! These kinds of estimates are used all the time. As I said, it is the same strategy underlying the [integral test](https://en.wikipedia.org/wiki/Integral_test_for_convergence). An even more general approach for estimating a sum is the [Euler-Maclaurin formula](https://en.wikipedia.org/wiki/Euler%E2%80%93Maclaurin_formula). – 2016-08-19