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How to find the value of $\lim_{n\to\infty}S(n)$, where $S(n)$ is given by $$S(n)=\displaystyle\sum_{k=1}^{n} \dfrac{k}{n^2+k^2}$$

Wolfram alpha is unable to calculate it.

This is a question from a questions booklet, and the options for the answer are--

$\begin{align} &A) \dfrac{\pi}{2} \\ &B) \log 2 \\ &C) \dfrac{\pi}{4} \\ &D) \dfrac{1}{2} \log 2 \end{align}$

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Hint: $S(n) = \frac{1}{n} \sum_{k=1}^{\infty} \frac{(k/n)}{1+(k/n)^2}.$ Think about Riemann sums. –  Ragib Zaman Jul 27 at 13:59
@RagibZaman We haven't yet been taught Riemann sums. –  Pkwssis Jul 27 at 13:59
@Pkwssis he means turn it into an integral. –  lemon Jul 27 at 14:00
@ozo We haven't been taught that either, as of yet. –  Pkwssis Jul 27 at 14:02
@DanielFischer No..it is k.. –  Pkwssis Jul 27 at 14:15

3 Answers 3

up vote 17 down vote accepted

Clearly, \begin{align} \lim_{n\to\infty}\sum_{k=1}^n\frac{k}{n^2+k^2} &=\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^n\frac{\frac{k}{n}}{1+\frac{k^2}{n^2}} \stackrel{\text{Riemann sum}}\longrightarrow \int_0^1 \frac{x\,dx}{1+x^2}=\left.\frac{1}{2}\log (1+x^2)\right|_0^1\\ &=\frac{1}{2}\log 2. \end{align}

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You could get away with a simple estimate of the expected range of the limit:

With $S_1(n)=\sum_{k=1}^{n}\frac{k}{n^2+n^2}$ and $S_2(n)=\sum_{k=1}^{n}\frac{k}{n^2}$, we have $$S_1(n)<S(n)<S_2(n)$$ The limits for $S_1$ and $S_2$ are $$\lim_{n\rightarrow\infty}S_1(n)=\lim_{n\rightarrow\infty}\frac{1}{2n^2}\sum_{k=1}^nk=\lim_{n\rightarrow\infty}\frac{1}{2n^2}\frac{n^2-n}{2}=\frac{1}{4}$$ $$\lim_{n\rightarrow\infty}S_2(n)=\lim_{n\rightarrow\infty}\frac{1}{n^2}\sum_{k=1}^nk=\lim_{n\rightarrow\infty}\frac{1}{n^2}\frac{n^2-n}{2}=\frac{1}{2}$$ So we know that $$\frac{1}{4}<\lim_{n\rightarrow\infty}S(n)<\frac{1}{2}.$$ Out of the given answers, that leaves only $\frac{1}{2}\log2$.

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Also saw this before checking with the Reimann sum approach. Good example of where multiple choice rewards estimation rather than just going by the book. –  Keith Jul 28 at 1:40

The general term is equivalent to $\frac{1}{k}$, so I don't think this converges at all.

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Note: The sum's upper limit has been corrected from $\infty$ to $n$ by the original poster. –  mvw Jul 27 at 14:30

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