Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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--

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

Here is a photo..enter image description here

share|improve this question
4  
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
1  
@RagibZaman We haven't yet been taught Riemann sums. –  Pkwssis Jul 27 at 13:59
1  
@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}

share|improve this answer

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 $\frac{1}{2}\log2$.

share|improve this answer
    
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.

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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.