# Evaluating $\lim_{n\to \infty} \sum_{r=1}^{n}\frac{r}{n^2+n+r}$

The motive is to evaluate the following limit:

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

I wrote it as

$$\lim_{n\to \infty}\sum_{r=1}^{n}\frac{r/n}{1 + 1/n + r/n^2} \approx ^{?} \int_{0} ^{1} x \, dx = \frac{1}{2}$$

Now is this correct? Doesn't seem very correct to me. Thanks for your thoughts :)

Oliver Oloa gave a hint on Sandwich theorem but removed answer.

$$\sum_{r=1}^{n} \frac{r}{n^2 + n + n} \le \sum_{r=1}^{n} \frac{r}{n^2 + n + r} \le \sum_{r=1}^{n} \frac{r}{n^2 + n + 1}$$

Using this I think we get $1/2 \le L \le 1/2$ so limit is $1/2$.

• Oops I made a mistake in my answer... – Olivier Oloa Jan 1 '18 at 12:02
• You can convert into an integral only if you get something $(1/n)\sum f(r/n)$. You should first get a factor of $1/n$ outside the sum. I will give it a try and post an answer if I succeed. – Paramanand Singh Jan 1 '18 at 12:06
• After you last update the limit should be $1/2$ which matches your integral answer. – Paramanand Singh Jan 1 '18 at 12:09
• Just note that the sum can be written as $(1/n)\sum (r/n) (1+1/n+r/n^2)^{-1}=(1/n)\sum (r/n) +o(1)$ so that the integral also works fine. – Paramanand Singh Jan 1 '18 at 12:12
• @Paramanand Do we have rules about when this ignoring works and when it doesn't ? – samjoe Jan 1 '18 at 12:13

Hint. One may also write $$\sum_{r=1}^{n} \frac{r}{n^2 + n + r}=(n^2+n)\left(H_{n^2+n}-H_{n^2+2n}\right)+n$$ and conclude, as $n \to \infty$, with the asymptotics $$H_n=\ln n+\gamma+\frac1{2n}+O\left(\frac1{n^2}\right).$$

Oliver Oloa gave a hint on Sandwich theorem but removed answer.

$$\sum_{r=1}^{n} \frac{r}{n^2 + n + n} \le \sum_{r=1}^{n} \frac{r}{n^2 + n + r} \le \sum_{r=1}^{n} \frac{r}{n^2 + n + 1}$$

Using this we get:

$$\frac{n(n+1)}{2(n^2 + n + n)} \le \sum_{r=1}^{n} \frac{r}{n^2 + n + r} \le \frac{n(n+1)}{2(n^2 + n + 1)}$$

So as $n\to \infty$ the limit is $1/2$

• You should accept your answer above which is OK for future refence. Thank you. – Olivier Oloa Jan 1 '18 at 12:40

The limit is $\frac{1}{2}$, since $\sum_{r=1}^{n}r = \frac{1}{2}n^2+\frac{1}{2}n$ and both $n^2+2n$ and $n^2+n+1$ are $n^2+O(n)$.

• Ok Jack but there is error in the first method, what is that (answer is same by fluke) – samjoe Jan 1 '18 at 12:11
• @samjoe: the first method is not incorrect: the wanted sum is not a Riemann sum strictly speaking, but it can be approximated by Riemann sums. It is enough to justify $\approx$ properly. – Jack D'Aurizio Jan 1 '18 at 12:13
• Yes I mean how do we do that – samjoe Jan 1 '18 at 12:14
• @samjoe: by approximating $\frac{r}{n^2+n+r}$ with $\frac{r}{n^2}$ and $\frac{r}{(n+1)^2}$, for instance. – Jack D'Aurizio Jan 1 '18 at 12:16
• last comment was very useful for me! Thanks a lot :) – samjoe Jan 2 '18 at 14:32


Note that

$$0 < \sum_{r = 1}^{n}{nr + r^{2} \over \pars{n^{2} + n + r}n^{2}} < \sum_{r = 1}^{n}{2n^{2} \over \pars{n^{2} + n + 1}n^{2}} = {2n \over n^{2} + n + 1}\,\,\,\stackrel{\mrm{as}\ n\ \to\ \infty}{\to}\,\,\,{\large 0}$$ such that \begin{align} \lim_{n\to \infty}\sum_{r = 1}^{n}{r \over n^{2} + n + r} & = \lim_{n\to \infty}\pars{% {1 \over n}\sum_{r = 1}^{n}{r \over n}} = \int_{0}^{1}x\,\dd x = \bbx{1 \over 2} \end{align}