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Compute the given limit $$ \lim_{n \to \infty }\frac{981}{n+5}\sum_{i=1}^{n} \left (\frac{i^2}{n^2} \right) $$ The sum is:

Can someone please show me the steps to complete this problem? The answer I arrived at was 0 but the homework program is telling me that it's wrong.

Thank you.

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  • $\begingroup$ You should consider upvoting or accepting an answer (by clicking the $\checkmark$) if an answer helped you. $\endgroup$ – Eff Nov 26 '14 at 22:14
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Firstly calculate the sum$$\sum_{i=1}^{n}\frac{i^2}{n^2}=\frac{1}{n^2}\sum_{i=1}^{n}i^2=\frac{1}{n^2}\frac{n(n+1)(2n+1)}{6}=\frac{\not n^2}{\not{n^2}}\frac{(1+\frac1n)(2n+1)}{6}$$ Therefore $$\begin{align*}\lim_{n \to \infty}\frac{981}{n+5}\frac{(1+\frac1n)(2n+1)}{6}&=\lim_{n\to \infty}\frac{\not n}{\not n}\frac{981(1+\frac1n)(2+\frac1n)}{6(1+\frac5n)}\overset{\frac1n\to 0}=\\\\&=\frac{981(1+0)\cdot(2+0)}{6(1+0)}=\frac{981\cdot2}{6}=\frac{981}{3}=327\end{align*}$$

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  • $\begingroup$ Thank you. Can you please outline the formula behind doing this type of problem with the limit involved? I haven't learned anything about it in class or in my textbooks. edit: My homework program also says that your answer is incorrect. $\endgroup$ – user70019 Nov 26 '14 at 21:44
  • $\begingroup$ You are welcome. I edited my answer, is it better now? $\endgroup$ – Jimmy R. Nov 26 '14 at 21:47
  • $\begingroup$ Yes, this is great. I didn't notice that you changed 981 to 985. $\endgroup$ – user70019 Nov 26 '14 at 21:51
  • $\begingroup$ Wait, the 985 was a mistake! The 981 is the correct! $\endgroup$ – Jimmy R. Nov 26 '14 at 21:52
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    $\begingroup$ Hey, this is a great explanation. I was just able to do another one of these by hand. $\endgroup$ – user70019 Nov 26 '14 at 22:07
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Hint1: $$\begin{split} \lim_{n \to \infty }\frac{981}{n+5}\sum_{i=1}^{n} \left(\frac{i^2}{n^2} \right) &= \lim_{n \to \infty} \frac{981}{n+5} \cdot \frac{\sum_{i=1}^{n}i^2}{n^2}\\ \end{split}$$

Hint2:

$$ \sum_{i=1}^{n}i^2 = \frac{n\cdot (n+1) \cdot 2n+1)}{6}$$


$$\begin{split} \lim_{n \to \infty }\frac{981}{n+5}\sum_{i=1}^{n} \left(\frac{i^2}{n^2} \right) &= \lim_{n \to \infty} \frac{981}{n+5} \frac{\sum_{i=1}^{n}i^2}{n^2}\\ &= \lim_{n \to \infty}\frac{981}{n+5} \frac{n(n+1)(2n+1)}{6 \cdot n^2}\\ &= \lim_{n \to \infty} \frac{n}{n} \cdot \frac{981 (2n^2+3n+1)}{(6n^2 +30n)}\\ &= \lim_{n \to \infty}\frac{n^2}{n^2} \cdot \frac{981\cdot(2+\frac{3}{n} + \frac{1}{n^2})}{6(1 + \frac{30}{n})} \\ &= \lim_{n \to \infty}\frac{327\cdot(2+\frac{3}{n} + \frac{1}{n^2})}{2(1 + \frac{30}{n})} &= 327 \end{split}$$

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  • $\begingroup$ What do I do with the limit? The limit is 0, right? $\endgroup$ – user70019 Nov 26 '14 at 21:46
  • $\begingroup$ No, 327 as I wrote. Limit of denominator is $2$ and nominator $327 \cdot 2$. Look, $\lim_{n\to\infty}\frac{3}{n} = 0$, $\lim_{n\to\infty}\frac{1}{n^2} = 0$, so $\lim_{n\to\infty}\left(2 + \frac{3}{n} + \frac{1}{n^2}\right) = 2$. And so on. $\endgroup$ – Tacet Nov 26 '14 at 21:51
  • $\begingroup$ Where does the following come from: $$ \sum\limits_{i=1}^n i^2=\frac{n\left(n+1\right)\left(2n+1\right)}{6} $$ $\endgroup$ – jm324354 Nov 26 '14 at 22:01
  • $\begingroup$ That's the summation formula for i^2. $\endgroup$ – user70019 Nov 26 '14 at 22:10
  • $\begingroup$ @bd1251252: It's well known fact. Exists so many derivations of this formula. You can find some in Internet, I believe, here too. An interesting way is to note the obvious fact $$ (a_2 - a_1) + (a_3 - a_2) + \dots + (a_{n+1} - a_{n}) = a_{n+1}-a_1$$ And let $ a_i = \frac{(2i - 1)^3}{24}$. I left you rest. :) $\endgroup$ – Tacet Nov 26 '14 at 22:18
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Hint: Pull the $\frac{1}{n^2}$ out of the sum and use the identity $$\sum_{i=1}^ni^2= \frac{n(n+1)(2n+1)}{6}$$

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I see a (calculus) tag. So ... that homework problem must follow a chapter on Riemann sums ... Let's try that.

Consider the integral $$ \int_0^1 x^2\,dx = \frac{1}{3} . $$ That integrand $x^2$ is continuous, therefore Riemann integrable. Consider the "upper sums" for $n$ equal-size intervals.

$$ \sum_{i=1}^n \left(\frac{i}{n}\right)^2\cdot\frac{1}{n} \to \frac{1}{3} $$ as $n \to \infty$. So the original probem must have answer $981/3 = 327$.

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$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ With Stolz-Ces$\grave{a}$ro Theorem: \begin{align} \color{#f00}{\lim_{n \to \infty}{981 \over n + 5} \sum_{i = 1}^{n}\pars{i^{2} \over n^{2}}} & = 981\lim_{n \to \infty}{1 \over \pars{n + 6}\pars{n + 1}^{2} - \pars{n + 5}n^{2}} \pars{\sum_{i = 1}^{n + 1}i^{2} - \sum_{i = 1}^{n}i^{2}} \\[3mm] &= 981\lim_{n \to \infty}{n^{2} + 2n + 1 \over 3n^{2} + 13n + 6} = 327\lim_{n \to \infty}{1 + 2/n + 1/n^{2} \over 1 + 13/\pars{3n} + 2/n^{2}} = \color{#f00}{327} \end{align}

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