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Using Riemann integrals of suitably functions, find the following limit

$$\lim_{n\to \infty}\sum_{k=1}^n \frac{k}{n^2+k^2}$$

Please help me check my method:

Let $$f(x)=\frac{x}{1+x^2}$$ For each n$\in$ $\Bbb N$, let partition $$P_n=({\frac{k}{n}:0\le k\le n})$$ and $$\xi^{(n)}=(\frac{1}{n},\frac{2}{n},...,\frac{n-1}{2n},1)$$ and $||P_n||=\frac{1}{n} \rightarrow 0$

$$\lim_{n\to \infty}\sum_{k=1}^n \frac{k}{n^2+k^2}=\int_0^1 \frac{x}{1+x^2}dx=\frac{1}{2} \ln(1+x^2)|^1_0=\frac{\ln 2}{2}$$

Is there any other methods for this question?

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marked as duplicate by YuiTo Cheng, Lord Shark the Unknown, Thomas Shelby, cmk, José Carlos Santos limits Jun 29 at 17:16

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    $\begingroup$ That seems correct to me. I personally can't think of another method... $\endgroup$ – Albert Zhang Oct 13 '13 at 4:09
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    $\begingroup$ What @AlbertZhang said. You might want to add a line where the sum is shown to be $\frac1n\sum\limits_{k=1}^n\frac{(k/n)}{1+(k/n)^2}$. $\endgroup$ – Did Oct 13 '13 at 10:41
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Your approach is fine and Riemann sums are definitely the way to go.

Anyway, I will show you an interesting overkill. Since: $$ \frac{2k}{k^2+n^2}=\frac{1}{k+in}+\frac{1}{k-in}=\int_{0}^{+\infty}e^{-kx}\left(e^{-inx}+e^{inx}\right)\,dx $$ we may write the original sum as: $$\begin{eqnarray*} S_n=\sum_{k=1}^{n}\int_{0}^{+\infty}\cos(nx) e^{-kx}\,dx &=& \int_{0}^{+\infty}\frac{1-e^{-nx}}{e^x-1}\cos(nx)\,dx\\&=&\int_{0}^{+\infty}\frac{(1-e^{-x})\cos x}{n(e^{x/n}-1)}\,dx\\&=&\int_{0}^{+\infty}\frac{\cos x-e^{-x}}{n(e^{x/n}-1)}\,dx+\int_{0}^{+\infty}\frac{1-\cos x}{e^x n (e^{x/n}-1)}\,dx.\end{eqnarray*}$$

Now you may notice that $n(e^{x/n}-1)$ is pointwise convergent to $x$ as $n\to +\infty$ and check that: $$ \int_{0}^{+\infty}\frac{\cos x-e^{-x}}{x}\,dx = 0. $$

So, by the dominated convergence theorem we have $$ \lim_{n\to +\infty} S_n = \int_{0}^{+\infty}\frac{1-\cos x}{xe^{x}}\,dx = \text{Re}\log(1+i) = \log\|1+i\| = \log\sqrt{2} = \color{red}{\frac{\log 2}{2}}$$ through the Cantarini-Frullani's theorem.

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