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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 $1/2$.

jonsno
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  • Oops I made a mistake in my answer... – Olivier Oloa Jan 01 '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 01 '18 at 12:06
  • After you last update the limit should be $1/2$ which matches your integral answer. – Paramanand Singh Jan 01 '18 at 12:09
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    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 01 '18 at 12:12
  • @Paramanand Do we have rules about when this ignoring works and when it doesn't ? – jonsno Jan 01 '18 at 12:13
  • Oh i have just written it as $o(1)$ but a full solution needs to prove this. You can expand $(1+1/n+r/n^2)^{-1}$ using binomial or perhaps give an estimate using inequalities. – Paramanand Singh Jan 01 '18 at 12:14
  • @Paramanand Sorry that question is very useful but I had one more question about it which was not addressed there ! – jonsno Jan 01 '18 at 12:22
  • I think Jack D'Aurizio has already addressed the problem of fixing the integral approach using the estimates. You should try based on his comment. – Paramanand Singh Jan 01 '18 at 12:24

4 Answers4

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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 $1/2$

jonsno
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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 $$ H_n=\ln n+\gamma+\frac1{2n}+O\left(\frac1{n^2}\right). $$

Olivier Oloa
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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 $n^2+O(n)$.

Jack D'Aurizio
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1

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \lim_{n\to \infty}\sum_{r = 1}^{n}{r \over n^{2} + n + r} & = \lim_{n\to \infty}\bracks{% {1 \over n}\sum_{r = 1}^{n}{r \over n} + \sum_{r = 1}^{n}\pars{{r \over n^{2} + n + r} - {r \over n^{2}}}} \\[5mm] & = \lim_{n\to \infty}\bracks{% {1 \over n}\sum_{r = 1}^{n}{r \over n} - \sum_{r = 1}^{n}{nr + r^{2} \over \pars{n^{2} + n + r}n^{2}}} \end{align}

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}

Felix Marin
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