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Considering the sum as a Riemann sum, evaluate $$\lim_{n\to\infty}\sum_{k=1}^{n}\frac{k}{n^2+k^2} .$$

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marked as duplicate by YuiTo Cheng, mrtaurho, Thomas Shelby, cmk, The Count Jun 30 at 0:00

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  • $\begingroup$ Hint: Pull out a factor of $\frac{1}{n}$ from the summand. $\endgroup$ – Ragib Zaman Dec 12 '12 at 17:41
  • $\begingroup$ @RagibZaman how? $\endgroup$ – user315 Dec 12 '12 at 17:44
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    $\begingroup$ $\dfrac{1}{n}\dfrac{\frac{k}{n}}{1+\left(\frac{k}{n}\right)^2}$ $\endgroup$ – André Nicolas Dec 12 '12 at 17:47
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$$\sum_{k=1}^n\frac{k}{n^2+k^2}=\frac{1}{n^2}\sum_{k=1}^n\frac{k}{1+\left(\frac{k}{n}\right)^2}=\frac{1}{n}\sum_{k=1}^n\frac{\frac{k}{n}}{1+\left(\frac{k}{n}\right)^2}\xrightarrow [n\to\infty]{}\int_0^1\frac{x}{1+x^2}\,dx=\ldots$$

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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<x_1<...<\frac{i}{n}<...<x_n=1 \right\}\end{equation} Now we must choose our function $f:[0,1]\to \mathbb{R}$ wisely so that $$U_{f,\mathcal{P}}=\sum_{k=1}^n\frac{k}{n^2+k^2}$$ But \begin{equation}U_{f,\mathcal{P}}=\sum\limits_{i=1}^{n}{M_i(f)\left( x_i-x_{i-1} \right)}=\sum\limits_{i=1}^{n}\frac{\sup_{x\in [x_{{i-1}},x_i]}f(x)}n \end{equation} If we choose an increasing function this simplifies to $$\sum_{k=1}^n\frac{k}{n^2+k^2}=\sum_{k=1}^{n}\frac{f(x_k)}n$$ Matching the terms gives $$f(x_k)=\frac{kn}{n^2+k^2}\Rightarrow f(x)=\frac{xn^2}{n^2+n^2x^2}=\frac{x}{1+x^2}$$ That's how you can come up with $f$. The rest can be found in Antonio's answer

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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 $S_n\to\frac{1}{2}\log 2$.

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  • $\begingroup$ Really cool approach, I have never seen anything like that. How have you guessed that it will work? $\endgroup$ – Santiago Aug 19 '16 at 21:21
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    $\begingroup$ @Santiago: Thanks! These kinds of estimates are used all the time. As I said, it is the same strategy underlying the integral test. An even more general approach for estimating a sum is the Euler-Maclaurin formula. $\endgroup$ – symplectomorphic Aug 19 '16 at 21:24

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