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$$ \lim_{n\rightarrow \infty} \sum_{r= 0}^{n} \frac{r}{n^2+ r} $$

My attempt
Divide Nr and Dr by $n^2$
$$ \lim_{n \rightarrow \infty} \sum_{r= 0}^{n} \frac{r/n^2}{1+ r/n^2} $$ =0

Is it correct or not

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  • $\begingroup$ Compare this sum with $\sum r/n^2$. $\endgroup$
    – Paramanand Singh
    Feb 11, 2020 at 4:29
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    $\begingroup$ Hint: For all $r$ in this range we have $$\frac r{(n+1)^2}\le\frac r{n^2+r}\le\frac r{n^2}.$$ $\endgroup$ Feb 11, 2020 at 4:29
  • $\begingroup$ This is solved in AOPS. Haven't found it on our site yet. This is close (exact same estimates work). $\endgroup$ Feb 11, 2020 at 4:34
  • $\begingroup$ @metamorphy Paramanand Singh showed how to do that :-) $\endgroup$ Feb 11, 2020 at 4:49
  • $\begingroup$ Why there is downvote to my question $\endgroup$ Feb 11, 2020 at 5:05

4 Answers 4

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No it's not as n start to tends infinity the sum does not remain negligible. We can use sandwich theorm that is

if f(x)$\le$h(x)$\le$g(x) and $lim_{x\rightarrow c}$ g(x) =$lim_{x\rightarrow c}$ f(x) = l then $lim_{x\rightarrow c}$ h(x)=l

For $\lim_{n\rightarrow \infty}$

$\sum_{r=0}^{\infty}\frac{r}{n^2}$ $\ge$ $\sum_{r=0}^{\infty}\frac{r}{n^2+r}$ $\ge$ $\sum_{r=0}^{\infty}\frac{r}{n^2+n}$

Therefore

$\frac{n(n+1)}{2n^2}$ $\ge$ $\sum_{r=0}^{\infty}\frac{r}{n^2 + r}$ $\ge$ $\frac{n(n+1)}{2n(n+1)}$

$\frac{1}{2}$ $\ge$ $\sum_{r=0}^{\infty}\frac{r}{n^2+r }$ $\ge$ $\frac{1}{2}.$

Hence answer is $\frac{1}{2}$

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  • $\begingroup$ You should fix your latex. Otherwise things are OK and +1 for that. $\endgroup$
    – Paramanand Singh
    Feb 11, 2020 at 4:30
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Here is one approach based on Riemann sum. The expression under limit can be written as $$\frac{1}{n}\sum_{r=1}^{n}\frac{rn}{n^2+r}$$ and note that $$t_r=\frac{nr} {n^2+r}\in\left[\frac{r-1}{n},\frac{r}{n}\right]$$ and hence the sum under limit is a Riemann sum for partition $$\{0,1/n,2/n,\dots,(n-1)/n,1\}$$ of $[0,1]$ with tag points $t_r$ and function $f(x) =x$. The desired limit is then equal to $\int_{0}^{1}x\,dx=1/2$.

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  • $\begingroup$ LOL. You don't need to evaluate a function at the end points of the subintervals to get a Riemann sum. $\endgroup$ Feb 11, 2020 at 4:45
  • $\begingroup$ @JyrkiLahtonen: yes, end points are most common, but definition of Riemann sum provides for a more general approach. $\endgroup$
    – Paramanand Singh
    Feb 11, 2020 at 4:46
  • $\begingroup$ @Paramanand Singh can solve using that method were we change $1/n $ to $ dx$ and $r/n$ to $ x$. $\endgroup$ Feb 11, 2020 at 5:04
  • $\begingroup$ @AbhishekKumar: it's best not to think in that manner. Better learn definition of a Riemann sum and try to see if your sum under limit is in that form or not. $\endgroup$
    – Paramanand Singh
    Feb 11, 2020 at 5:13
  • $\begingroup$ @AbhishekKumar: see math.stackexchange.com/a/3522906/72031 $\endgroup$
    – Paramanand Singh
    Feb 11, 2020 at 5:15
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The limit and the summation cannot be exchanged. For example, Dominated Convergence does not help: consider the maximum for each $r$: $$ \sup_{n\ge r}\frac{r}{n^2+r}=\frac1{r+1} $$ since $\sum\limits_{r=1}^\infty\frac1{r+1}=\infty$, any dominating series diverges.


In fact $$ \begin{align} \sum_{r=1}^n\frac{r}{n^2+r} &\ge\sum_{r=1}^n\frac{r}{n^2+n}\\ &=\frac12 \end{align} $$ and $$ \begin{align} \sum_{r=1}^n\frac{r}{n^2+r} &\le\sum_{r=1}^n\frac{r}{n^2}\\ &=\frac12\left(1+\frac1n\right) \end{align} $$ The Squeeze Theorem then says $$ \lim_{n\to\infty}\sum_{r=1}^n\frac{r}{n^2+r}=\frac12 $$

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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[#ffe,10px]{\sum_{r = 0}^{n}{r \over n^{2} + r}} = n + 1 - n^{2}\sum_{r = 0}^{n}{1 \over r + n^{2}} \\[5mm] = &\ n + 1 - n^{2}\sum_{r = 0}^{\infty}\pars{{1 \over r + n^{2}} - {1 \over r + n + 1 + n^{2}}} \\[5mm] = &\ n + 1 - n^{2}\bracks{\Psi\pars{n^{2} + n + 1} - \Psi\pars{n^{2}}} \\[1cm] \stackrel{\mrm{as}\ n\ \to\ \infty}{\sim}\,\,\,& n + 1 \\[2mm] &\ \!\!\!\!\!\!\!\!\!\!\!\!\!\!\! -n^{2}\braces{\bracks{\ln\pars{n^{2} + n + 1} - {1 \over 2n^{2} + 2n + 2}} - \bracks{\ln\pars{n^{2}} - {1 \over 2n^{2}}}} \\[1cm] = &\ n + 1 - n^{2}\ln\pars{1 + {1 \over n} + {1 \over n^{2}}} + {n^{2} \over 2n^{2} + 2n + 2} - {1 \over 2} \\[5mm] \stackrel{\mrm{as}\ n\ \to\ \infty}{\sim}\,\,\,&\ n + 1 - n^{2}\bracks{\pars{{1 \over n} + {1 \over n^{2}}} - {1 \over 2}\pars{{1 \over n} + {1 \over n^{2}}}^{2}} \\[5mm] \stackrel{\mrm{as}\ n\ \to\ \infty}{\sim}\,\,\,&\ n + 1 - \bracks{n + 1 - n^{2}\,{1 \over 2}\pars{1 \over n}^{2}} \,\,\,\stackrel{\mrm{as}\ n\ \to\ \infty}{\to}\,\,\, \bbx{\large{1 \over 2}} \end{align}

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