17
$\begingroup$

Evaluate the limit

$$\lim_{n \to \infty} \dfrac{1}{1+n^2} +\dfrac{2}{2+n^2}+ \ldots+\dfrac{n}{n+n^2}$$

My approach :

If I divide numerator and denominator by $n^2$ I get :

$$\lim_{ n \to \infty} \dfrac{\frac{1}{n^2}}{\frac{1}{n^2} +1} +\dfrac{\frac{2}{n^2}}{\frac{2}{n^2} +1} + \ldots+ \dfrac{\frac{1}{n}}{\frac{1}{n} + 1}=0$$

but the answer is $\dfrac{1}{2}$ please suggest how to solve this.. thanks.

$\endgroup$
2
  • 1
    $\begingroup$ Do you familiar with Riemann's sums? $\endgroup$ Commented Dec 24, 2013 at 16:15
  • 6
    $\begingroup$ And what is $\lim_{n\to\infty}\frac{1}{n}+\frac{1}{n}+\cdots+ \frac{1}{n}$, where $\frac{1}{n}$ appears $n$ times? $\endgroup$ Commented Dec 24, 2013 at 16:17

5 Answers 5

22
$\begingroup$

For each $1\leq i\leq n$, $\frac{1}{n^2+i}\leq \frac{1}{n^2}$ and $\frac{1}{n^2+i}\geq \frac{1}{n^2+n}$, and so we may bound the sum from above and below by $$\sum_{i=1}^{n}\frac{i}{n+n^{2}}\leq\sum_{i=1}^{n}\frac{i}{i+n^{2}}\leq\sum_{i=1}^{n}\frac{i}{n^{2}}.$$ Since $\sum_{i=1}^{n}i=\frac{n(n+1)}{2},$ this becomes $$\frac{1}{2}=\frac{n(n+1)}{2n(n+1)}\leq\sum_{i=1}^{n}\frac{i}{i+n^{2}}\leq\frac{n(n+1)}{2n^{2}}=\frac{1}{2}\left(1+\frac{1}{n}\right),$$ and so it follows from the squeeze theorem that the limit is $\frac{1}{2}$.

$\endgroup$
3
  • $\begingroup$ brilliant! (all that follows is extra text inserted to satisfy the software censor's rule that "comments MUST BE at least 15 characters in length". I'm sure this excellent, and absolutely non-arbitrary rule must keep a lot of footling short comments from wasting certain peoples' very valuable time). I only wish there were far more such rules applied to comments, to encourage a more structured approach to the process of commenting. $\endgroup$ Commented Dec 24, 2013 at 19:20
  • $\begingroup$ Short and sweet! $\endgroup$ Commented Dec 27, 2013 at 2:13
  • $\begingroup$ @David, I believe the rule is to avoid myriad comments such as "me too" that will typically appear in any public forum, adding little or nothing to the information content and obscuring potentially more valuable contributions $\endgroup$
    – holdenweb
    Commented Oct 19, 2016 at 13:02
13
$\begingroup$

$$ S_n=\sum_{k=1}^n \frac{k}{k+n^2} = \sum_{k=1}^n \left( 1 -\frac{n^2}{k+n^2} \right) \\ = n - \sum_{k=1}^n \left(\frac{n^2}{k+n^2} \right) $$ but $$\sum_{k=1}^n \left(\frac{n^2}{k+n^2} \right) = \sum_{k=1}^n \frac1{1+\frac{k}{n^2}}$$

$$ = \sum_{k=1}^n \sum_{j=0}^{\infty} \left(\frac{-k}{n^2} \right)^j \\ = n -\frac1{n^2}\frac{n(n+1)}{2} +O\left(\frac1{n}\right) = n - \frac12 +O\left(\frac1{n}\right) $$ so $$ \lim_{n \rightarrow \infty} S_n = n - (n - \frac12) = \frac12 $$

$\endgroup$
0
4
$\begingroup$

Since $$ 1+n^2 \le k+n^2\le n+n^2 \quad \forall k \in \{1,2,\ldots,n\}, $$ it follows that $$ \frac{n(n+1)}{2(n+n^2)}=\sum_{k=1}^n\frac{k}{n+n^2}\le \sum_{k=1}^n\frac{k}{k+n^2}\le\sum_{k=1}^n\frac{k}{1+n^2}=\frac{n(n+1)}{2(1+n^2)} \quad \forall n \ge 1. $$ Thus $$ \lim_n\sum_{k=1}^n\frac{k}{k+n^2}=\frac12. $$

$\endgroup$
4
  • 2
    $\begingroup$ But, $$\lim_{n \to \infty} \frac1n\sum_{r=1}^n f\left(\frac rn\right)=\int_0^1f(x)dx$$ $\endgroup$ Commented Dec 24, 2013 at 16:24
  • 1
    $\begingroup$ This is just Eric's answer. $\endgroup$ Commented Dec 24, 2013 at 18:23
  • $\begingroup$ @lab (+1) thank you for that succinct statement. I have previously had to go through the angst of detail every time i have applied this line of reasoning. now I can just apply this result. $\endgroup$ Commented Dec 24, 2013 at 19:37
  • $\begingroup$ @labbhattacharjee This is not a Riemann sum. So, one would need to be careful in applying such a framework. $\endgroup$
    – Mark Viola
    Commented Feb 17, 2017 at 20:15
1
$\begingroup$

Use Riemann sums to show that $S\in\Big[\tfrac{\ln2}2,\tfrac12\Big]$ :

$$\sum_{k=1}^n\frac k{k^2+n^2}<\sum_{k=1}^n\frac k{k+n^2}<\sum_{k=1}^n\frac k{n^2}\quad\iff\quad\int_0^1\frac x{1+x^2}dx<S<\int_0^1xdx$$

$\endgroup$
3
  • $\begingroup$ This provides bounds, but does not yield the limit. Apology in advance if I've missed something here. -Mark $\endgroup$
    – Mark Viola
    Commented Feb 17, 2017 at 20:17
  • $\begingroup$ @Dr.MV: I was able to determine $\tfrac12$ as the upper limit, but was unable, at the time, to provide a better lower limit. Meanwhile, other answers, showing that the latter is also $\tfrac12,$ have already been posted. $\endgroup$
    – Lucian
    Commented Feb 17, 2017 at 22:03
  • $\begingroup$ No worry. Your answers are often bedazzling. So, I thought I might have missed something. -Mark $\endgroup$
    – Mark Viola
    Commented Feb 17, 2017 at 22:26
-1
$\begingroup$

Answer

$$\lim_{n \to \infty} \dfrac{1}{1+n^2} +\dfrac{2}{2+n^2}+ \ldots+\dfrac{n}{n+n^2}$$

$$\lim_{n \to \infty} \dfrac{1}{n^2}(\dfrac{1}{1/n^2 + 1} +\dfrac{2}{2/n^2+1}+ \ldots+\dfrac{n}{n/n^2+1})$$

$$\lim_{n \to \infty} \dfrac{1}{n^2}\dfrac{n*(n+1)}{2}$$

$$\lim_{n \to \infty}\dfrac{1}{2}*(1+\dfrac{1}{n})$$

$$=\dfrac {1}{2}$$

$\endgroup$
5
  • 5
    $\begingroup$ What happened from line 2 to 3? $\endgroup$ Commented Dec 24, 2013 at 16:43
  • 1
    $\begingroup$ maybe he or she use limit first as $1/n^2\to 0$, big mistake! $\endgroup$
    – mathlove
    Commented Dec 24, 2013 at 16:48
  • $\begingroup$ Can I not use product of limits? $\endgroup$ Commented Dec 24, 2013 at 16:51
  • 1
    $\begingroup$ Not the way you are doing it here. $\endgroup$ Commented Dec 24, 2013 at 17:02
  • $\begingroup$ Understood, David's method is what I intended but applied limits improperly $\endgroup$ Commented Dec 24, 2013 at 17:07

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .