Can't figure out this limit of a sum $\lim_{n\to\infty}\sum_{k=0}^{2n+1}\frac{n}{n^{2}+k}$

Question

so basically i should find the limit in (+infinity) of this sum $\sum_{k=0}^{2n+1}\frac{n}{n^{2}+k}$

I tried to simplify this sum in : $\sum_{k=0}^{2n+1}\frac{1}{n+\frac{k}{n}}$

and started from this equation : $0< k< 2n+1$

and stopped here : $\frac{1}{2+n+\frac{1}{n}}< \frac{1}{n+\frac{k}{n}}< \frac{1}{n}$

now i want to "sum" this inequality to have the first sum but i don't know how and thanks for replying .

If you sum $2n+1$ terms and every term is between $\frac{1}{n}$ and $\frac{1}{n+1}$, their sum cannot differ from $2$ by much. — Jack D'Aurizio, Dec 28 '16 at 22:43
i ve made a typo mistake let me please rework that inequality — Ayman Erroutabi, Dec 28 '16 at 22:54
Perhaps looking at similar past questions might help you: Limit $\lim_{n\to\infty}\sum_{k=0}^n\frac{n}{n^2+k}$. Found using Approach0. — Martin Sleziak, Dec 29 '16 at 08:55
Please, try to make the titles of your questions more informative. E.g., Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here. — Martin Sleziak, Dec 29 '16 at 08:55

score 3 · Accepted Answer · edited Dec 28 '16 at 23:15

3

With all the steps: if $k$ is in the range $[0,2n+1]$, then $\frac{n}{n^2+k}$ is in the range $\left[\frac{n}{(n+1)^2},\frac{n}{n^2}\right]$, so: $$ \frac{n(2n+1)}{(n+1)^2}\leq \sum_{k=0}^{2n+1}\frac{n}{n^2+k}\leq \frac{n(2n+1)}{n^2} $$ and squeezing applies, since the limit as $n\to \infty$ of both the RHS and the LHS is just $\color{red}{\large 2}$.

edited Dec 28 '16 at 23:15

Ayman Erroutabi

361

answered Dec 28 '16 at 22:56

Jack D'Aurizio

353,855

Ah, beautifully simple. – Simply Beautiful Art Dec 28 '16 at 22:57
so by multiplying both extremes with final value ($2n+1$) we get the sum in middle ? thats new . thank you man – Ayman Erroutabi Dec 28 '16 at 23:04
@AymanErroutabi Very elementary too, I like it much. – Simply Beautiful Art Dec 28 '16 at 23:05
@Jack D'Aurizio Thanks man for ur time and efforts i appreciate that . – Ayman Erroutabi Dec 28 '16 at 23:13

Claude Leibovici · Answer 2 · 2016-12-29T09:00:49.497

If you use harmonic numbers $$S_n=\sum_{k=0}^{2n+1}\frac n{n^2+k}=n\sum_{k=0}^{2n+1}\frac 1{n^2+k}=n \left(H_{(n+1)^2}-H_{(n^2-1)}\right)$$ Now, using for large values of $p$ the asymptotics $$H_p=\gamma +\log \left(p\right)+\frac{1}{2 p}-\frac{1}{12 p^2}+O\left(\frac{1}{p^3}\right)$$ apply to each term and continue Taylor expansions to get $$H_{(n+1)^2}-H_{(n^2-1)}=\frac{2}{n}-\frac{1}{3 n^3}+O\left(\frac{1}{n^4}\right)$$ $$S_n=2-\frac{1}{3 n^2}+O\left(\frac{1}{n^3}\right)$$ which shows the limit and how it is approached.

In the same spirit, we could show that $$T_n=\sum_{k=0}^{an+b}\frac n{n^2+k}=a+\frac{b-\frac{a^2}{2}+1}{n}+\frac{a \left(2 a^2-6 b-3\right)}{6 n^2}+O\left(\frac{1}{n^3}\right)$$

Simply Beautiful Art · Answer 3 · 2016-12-28T23:03:45.433

1

According to the integral test,

$$\int_0^{2n+2}\frac n{n^2+x}\ dx\le\sum_{k=0}^{2n+1}\frac n{n^2+k}\le\frac1n+\int_0^{2n+1}\frac n{n^2+x}\ dx$$

which reduces down to

$$n\ln\left(\frac{n^2+2n+2}{n^2}\right)\le\sum_{k=0}^{2n+1}\frac n{n^2+k}\le\frac1n+2n\ln\left(\frac{n+1}n\right)$$

Thus, by the squeeze theorem,

$$\lim_{n\to\infty}\sum_{k=0}^{2n+1}\frac n{n^2+k}=2$$

edited Dec 28 '16 at 23:03

answered Dec 28 '16 at 22:52

Simply Beautiful Art

74,685

sorry to break your efforts guys but i forgot that k=0 .. sorryyyy – Ayman Erroutabi Dec 28 '16 at 22:59
@AymanErroutabi That's easy to fix, no worries. – Simply Beautiful Art Dec 28 '16 at 23:00
@AymanErroutabi Hahaha, if I may point out, when $k=0$ and $n\to\infty$, we have $0$, so it has no affect. – Simply Beautiful Art Dec 28 '16 at 23:04
so you mean the limit would equal 0 at the end ? – Ayman Erroutabi Dec 28 '16 at 23:07
@AymanErroutabi at $k=0$ the limit is $0$, so $k=0$ does not affect the limit. – Simply Beautiful Art Dec 28 '16 at 23:08
oh my bad my bad . i just tried it on paper it equals 2 thank you guys for the support ^^ – Ayman Erroutabi Dec 28 '16 at 23:09
@AymanErroutabi Give the other guy your sincerity, he deserves most of it. – Simply Beautiful Art Dec 28 '16 at 23:10

Can't figure out this limit of a sum $\lim_{n\to\infty}\sum_{k=0}^{2n+1}\frac{n}{n^{2}+k}$

3 Answers3