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.

  • 1
    $\begingroup$ Do you familiar with Riemann's sums? $\endgroup$ – Salech Rubenstein Dec 24 '13 at 16:15
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    $\begingroup$ And what is $\lim_{n\to\infty}\frac{1}{n}+\frac{1}{n}+\cdots+ \frac{1}{n}$, where $\frac{1}{n}$ appears $n$ times? $\endgroup$ – Salech Rubenstein Dec 24 '13 at 16:17

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 $\frac{1}{2}$.

  • $\begingroup$ 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. $\endgroup$ – David Holden Dec 24 '13 at 19:20
  • $\begingroup$ Short and sweet! $\endgroup$ – ireallydonknow Dec 27 '13 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 Oct 19 '16 at 13:02

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


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 $$ \lim_n\sum_{k=1}^n\frac{k}{k+n^2}=\frac12. $$

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    $\begingroup$ But, $$\lim_{n \to \infty} \frac1n\sum_{r=1}^n f\left(\frac rn\right)=\int_0^1f(x)dx$$ $\endgroup$ – lab bhattacharjee Dec 24 '13 at 16:24
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    $\begingroup$ This is just Eric's answer. $\endgroup$ – Andrés E. Caicedo Dec 24 '13 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. $\endgroup$ – David Holden Dec 24 '13 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 Feb 17 '17 at 20:15

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

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

  • $\begingroup$ This provides bounds, but does not yield the limit. Apology in advance if I've missed something here. -Mark $\endgroup$ – Mark Viola Feb 17 '17 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 Feb 17 '17 at 22:03
  • $\begingroup$ No worry. Your answers are often bedazzling. So, I thought I might have missed something. -Mark $\endgroup$ – Mark Viola Feb 17 '17 at 22:26


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

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

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

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

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

  • 5
    $\begingroup$ What happened from line 2 to 3? $\endgroup$ – Andrés E. Caicedo Dec 24 '13 at 16:43
  • 1
    $\begingroup$ maybe he or she use limit first as $1/n^2\to 0$, big mistake! $\endgroup$ – mathlove Dec 24 '13 at 16:48
  • $\begingroup$ Can I not use product of limits? $\endgroup$ – Satish Ramanathan Dec 24 '13 at 16:51
  • 1
    $\begingroup$ Not the way you are doing it here. $\endgroup$ – Andrés E. Caicedo Dec 24 '13 at 17:02
  • $\begingroup$ Understood, David's method is what I intended but applied limits improperly $\endgroup$ – Satish Ramanathan Dec 24 '13 at 17:07

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