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What ways would you propose for the limit below? $$\lim_{n\to\infty}n\sum_{k=1}^n\left( \frac{1}{(2k-1)^2} - \frac{3}{4k^2}\right)$$

Thanks in advance for your suggestions, hints!

Sis.

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    $\begingroup$ Boy, I went to the wrong high school! $\endgroup$
    – icurays1
    Feb 18 '13 at 14:48
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    $\begingroup$ My next question on the Riemann Hypothesis will be called "elementary school question". $\endgroup$
    – GEdgar
    Feb 18 '13 at 14:53
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    $\begingroup$ Where is this high school? $\endgroup$ Feb 18 '13 at 14:58
  • $\begingroup$ @Chris'ssisterandpals: I see. $\endgroup$ Feb 18 '13 at 15:02
  • $\begingroup$ I feel stupid now. I only got to this stuff at university level. $\endgroup$
    – Thomas
    Feb 19 '13 at 3:25
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Here is a high school level answer: $$ \begin{align} \sum_{k=1}^n\left(\frac1{(2k-1)^2}-\frac3{4k^2}\right) &=\sum_{k=1}^n\left(\frac1{(2k-1)^2}+\frac1{(2k)^2}-\frac1{k^2}\right)\\ &=\sum_{k=1}^{2n}\frac1{k^2}-\sum_{k=1}^n\frac1{k^2}\\ &=\sum_{k=n+1}^{2n}\frac1{k^2}\tag{1} \end{align} $$ Using partial fractions and summing the telescoping series, we get $$ \hspace{-1cm} \frac1{n+1}-\frac1{2n+2} =\sum_{k=n+1}^{2n}\frac1{k(k+1)} \le\sum_{k=n+1}^{2n}\frac1{k^2} \le\sum_{k=n+1}^{2n}\frac1{k(k-1)} =\frac1n-\frac1{2n}\tag{2} $$ Therefore, the Squeeze Theorem and $(2)$ yield $$ \lim_{n\to\infty}n\sum_{k=n+1}^{2n}\frac1{k^2}=\frac12\tag{3} $$

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  • $\begingroup$ Oh, this is a candidate for the penultimate high school year since you used no integral. Just awesome! (+1) Actually, excepting the limit part, it could also be a good problem for the middle school. $\endgroup$ Feb 18 '13 at 17:08
  • $\begingroup$ @Chris'ssisterandpals: I don't usually post two answers to the same question, but the prerequisite level of these answers was so disparate, that I felt I should. $\endgroup$
    – robjohn
    Feb 18 '13 at 17:20
  • $\begingroup$ I have no problem with that. For me the answers are important! :-) $\endgroup$ Feb 18 '13 at 17:22
  • $\begingroup$ Very nice!${}{}$ $\endgroup$
    – mrf
    Feb 19 '13 at 11:56
  • $\begingroup$ (+1) The easier, the better. $\endgroup$ Sep 13 '14 at 18:01
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OK, it turns out that

$$\sum_{k=1}^n\left( \frac{1}{(2k-1)^2} - \frac{3}{4k^2}\right) = \sum_{k=1}^{n-1} \frac{1}{(k+n)^2}$$

This may be shown by observing that

$$\sum_{k=1}^n \frac{1}{(2k-1)^2} = \sum_{k=1}^{2 n-1} \frac{1}{k^2} - \frac{1}{2^2} \sum_{k=1}^n \frac{1}{k^2}$$

The desired limit may then be rewritten as

$$\lim_{n \rightarrow \infty} \frac{1}{n} \sum_{k=1}^{n-1} \frac{1}{(1 + (k/n))^2}$$

which you may recognize as a Riemann sum, equal to

$$\int_0^1 dx \: \frac{1}{(1+x)^2} = \frac{1}{2}$$

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  • $\begingroup$ nice work and well shown. (+1) $\endgroup$ Feb 18 '13 at 15:18
  • $\begingroup$ This is the way I wanted to go, but I went with the Euler-Maclaurin Sum Formula as an alternative :-) (+1) $\endgroup$
    – robjohn
    Feb 18 '13 at 16:46
  • $\begingroup$ Nice to see it done different ways. $\endgroup$
    – Ron Gordon
    Feb 18 '13 at 16:56
  • $\begingroup$ I have added yet another way :-) $\endgroup$
    – robjohn
    Feb 18 '13 at 17:15
  • $\begingroup$ You're an animal, @robjohn! $\endgroup$
    – Ron Gordon
    Feb 18 '13 at 20:09
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$$n\sum_{k=1}^n\frac{1}{(2k-1)^2}-\frac{3}{4k^2} =n(H_{2n}^{(2)}-H_{n}^{(2)})=\sum_{j=1}^n\frac{n}{(j+n)^2}\to\int_0^1\frac{dx}{(1+x)^2} =\frac{1}{2}$$

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  • $\begingroup$ This is incorrect. The limit indeed converges, but not to the limit you stated. $\endgroup$
    – nbubis
    Feb 18 '13 at 14:58
  • $\begingroup$ Could you tell me where I made a mistake? $\endgroup$ Feb 18 '13 at 15:01
  • $\begingroup$ I think he is right. $\endgroup$
    – Yimin
    Feb 18 '13 at 15:13
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    $\begingroup$ @Chis's sister and pals Where do you get these questions? $\endgroup$ Feb 18 '13 at 15:20
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    $\begingroup$ @IshanBanerjee: I have them from my brother. $\endgroup$ Feb 18 '13 at 16:12
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Using the Euler-Maclaurin Sum Formula, $$ \begin{align} \sum_{k=1}^n\left(\frac1{(2k-1)^2}-\frac3{4k^2}\right) &=\sum_{k=1}^n\left(\frac1{(2k-1)^2}+\frac1{(2k)^2}-\frac1{k^2}\right)\\ &=\sum_{k=1}^{2n}\frac1{k^2}-\sum_{k=1}^n\frac1{k^2}\\ &=\left(C-\frac1{2n}+O\left(\frac1{n^2}\right)\right)-\left(C-\frac1n+O\left(\frac1{n^2}\right)\right)\\ &=\frac1{2n}+O\left(\frac1{n^2}\right) \end{align} $$ Therefore, $$ \begin{align} \lim_{n\to\infty}n\sum_{k=1}^n\left(\frac1{(2k-1)^2}-\frac3{4k^2}\right) &=\lim_{n\to\infty}n\left(\frac1{2n}+O\left(\frac1{n^2}\right)\right)\\ &=\frac12 \end{align} $$

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  • $\begingroup$ Another good shot with Euler-Maclaurin Sum Formula. Thanks John! (+1) $\endgroup$ Feb 18 '13 at 16:51
  • $\begingroup$ @Chris'ssisterandpals: I am writing up a high school level answer now. $\endgroup$
    – robjohn
    Feb 18 '13 at 16:52
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$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{\lim_{n\ \to\ \infty}\bracks{n\sum_{k = 1}^{n}\fermi\pars{k}}:\ {\large ?}. \qquad\fermi\pars{k} \equiv {1 \over \pars{2k -1}^{2}} - {3 \over 4k^{2}}}$

Note that $\ds{\sum_{k = 1}^{\infty}\fermi\pars{k} = 0}$ because $$ \sum_{k = 1}^{\infty}\fermi\pars{k} =\sum_{k = 1}^{\infty}{1 \over \pars{2k - 1}^{2}} -\sum_{k = 1}^{\infty}{3 \over 4k^{2}} =\bracks{\sum_{k = 1}^{\infty}{1 \over k^{2}} -\sum_{k = 1}^{\infty}{1 \over \pars{2k}^{2}}} -\sum_{k = 1}^{\infty}{3 \over 4k^{2}} $$

With Stolz-Cesàro Theorem: \begin{align}&\color{#66f}{\large\lim_{n\ \to\ \infty}\bracks{% n\sum_{k = 1}^{n} \fermi\pars{k}}}=\lim_{n\ \to\ \infty}{\sum_{k = 1}^{n}% \fermi\pars{k} \over 1/n} =\lim_{n\ \to\ \infty}{% \sum_{k = 1}^{n + 1}\fermi\pars{k} - \sum_{k = 1}^{n}\fermi\pars{k} \over 1/\pars{n + 1} - 1/n} \\[3mm]&=\lim_{n\ \to\ \infty}\bracks{-n\pars{n + 1}\fermi\pars{n + 1}} =\lim_{n\ \to\ \infty}\braces{n\pars{n + 1} \bracks{{3 \over 4\pars{n + 1}^{2}} - {1 \over 4\pars{n + 1}^{2}}}} \\[3mm]&=\half\,\lim_{n\ \to\ \infty}{n \over n + 1}=\color{#66f}{\Large\half} \end{align}

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  • $\begingroup$ Good job there! (+1) $\endgroup$ Sep 13 '14 at 6:30
  • $\begingroup$ @Chris'ssis Thanks. You are always posting interesting questions. $\endgroup$ Sep 13 '14 at 6:40

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