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How to find $\lim_{n\to\infty}\left(\frac{\pi^2}{6}-\sum_{k=1}^n\frac{1}{k^2}\right)n$?

It is well-known that $\lim_{n\to\infty}\sum_{k=1}^n\frac{1}{k^2}=\frac{\pi^2}{6}$, so $$\lim_{n\to\infty}\left(\frac{\pi^2}{6}-\sum_{k=1}^n\frac{1}{k^2}\right)n=\lim_{n\to\infty}\frac{\frac{\pi^2}{6}-\sum_{k=1}^n\frac{1}{k^2}}{\frac{1}{n}}$$ has indeterminate form $\frac{0}{0}$.

But it can't use L'Hôpital's Rule.

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  • $\begingroup$ No l'Hospital because it's not a continuous function of $n$. Try to express it as $\sum_{k=n+1}^\infty$ and convert it into a Riemann sum. $\endgroup$
    – orion
    Aug 29, 2015 at 14:38
  • $\begingroup$ To start you want to notice that $\pi^2/6-\sum_{k=1}^n1/k^2=\sum_{k=n+1}^\infty1/k^2$. $\endgroup$
    – David C. Ullrich
    Aug 29, 2015 at 14:38
  • $\begingroup$ Would this help, $\frac{\pi^2}{6}-(\frac{\pi^2}{6}-\sum\limits^{\infty}_{i=k+1}\frac1{i^2})$?\ $\endgroup$
    – Aditya Agarwal
    Aug 29, 2015 at 14:39

4 Answers 4

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

This looks like a Riemann sum:

$$\lim_{n\to\infty}\frac{1}{n}\sum_{k=n+1}^\infty\frac{1}{(k/n)^2}=\int_{1}^\infty\frac{1}{x^2}dx=1$$ where $\frac{1}{n}$ plays the role of $dx$ in the limit and $x=k/n$.

If worried about the infinite upper limit, it's ok - because the integral of $1/x^2$ is absolutely convergent, the upper limit can be replaced by $k_{max}=Mn$ ($M>1$) and limited separately after the limit in $n$, and shown that it converges for any $M$.

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    $\begingroup$ Alternatively to your argument $$\frac{1}{n}\sum_{k=n+1}^\infty \frac{1}{(k/n)^2}\leq \int_{1}^{\infty}\frac{dx}{x^2}\leq \frac{1}{n}\sum_{k=n}^\infty \frac{1}{(k/n)^2} = \frac{1}{n} + \frac{1}{n}\sum_{k=n+1}^\infty \frac{1}{(k/n)^2}$$ Basically, you can compare $\frac{1}{x^2}$ with two different step functions. $\endgroup$
    – Thomas Andrews
    Aug 29, 2015 at 14:48
  • $\begingroup$ Good, sandwiching is a good proof. The point is, it works. $\endgroup$
    – orion
    Aug 29, 2015 at 14:49
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    $\begingroup$ Yes, and sandwich lets you know an upper bound on how fast it converges. $\endgroup$
    – Thomas Andrews
    Aug 29, 2015 at 14:50
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We may also use creative telescoping. Since: $$\left(\frac{1}{n}-\frac{1}{n+1}\right)\leq\frac{1}{n^2}\leq \left(\frac{1}{n-1}-\frac{1}{n}\right)\tag{1}$$ it follows that: $$ \frac{1}{n+1}\leq\sum_{k>n}\frac{1}{k^2}\leq \frac{1}{n}, \tag{2}$$ hence the limit is $\color{red}{1}$ by squeezing.

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  • $\begingroup$ Out of all the other answers here, this one is the simplest. +1 $\endgroup$
    – Paramanand Singh ♦
    Aug 30, 2015 at 2:17
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Alternatively, you can use the Stolz-Cesaro theorem, the counterpart of L'Hospital's rule for sequences. It yields that your limit is the same as $$\lim_{n\to\infty}\frac{1/(n+1)^2}{1/n - 1/(n+1)}=1. $$

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  • $\begingroup$ Stolz-Cesaro seems to want $a_n$ and $b_n$ divergent. The general form needs $b_n$ divergent. $\endgroup$
    – Thomas Andrews
    Aug 29, 2015 at 14:53
  • $\begingroup$ @ThomasAndrews: I hope I have read it works also when the sequences tend to $0$ (?) I must reckon it certainly wasn't Wikipedia, anyway... $\endgroup$
    – Vincenzo Oliva
    Aug 29, 2015 at 14:56
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    $\begingroup$ @ThomasAndrews: Found it: math.stackexchange.com/questions/599204/… $\endgroup$
    – Vincenzo Oliva
    Aug 29, 2015 at 14:56
  • $\begingroup$ Nice. So if $y_n$ is monotone and unbounded, and $x_n$ is a positive sequence with $\sum_1^{\infty} x_n = L$, then $$y_n(L-\sum_{k\leq n} x_k)$$ converges to the same value as $$\frac{y_{n+1}y_nx_{n+1}}{y_{n+1}-y_n}.$$ $\endgroup$
    – Thomas Andrews
    Aug 29, 2015 at 15:12
  • $\begingroup$ @ThomasAndrews: Precisely. When I learnt about Stolz-Cesaro I was amazed. I still am, often. $\endgroup$
    – Vincenzo Oliva
    Aug 29, 2015 at 15:26
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$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \lim_{n \to \infty}\bracks{\pars{{\pi^{2} \over 6} - \sum_{k = 1}^{n}{1 \over k^{2}}}n} &= \lim_{n \to \infty}\bracks{\pars{{1 \over n} - 2\int_{n}^{\infty}{\braces{x} \over x^{3}}\,\dd x}n}\label{1}\tag{1} \end{align}

where we used a well known identity which is related to the Riemann Zeta function.


Moreover, $\ds{0 < 2\int_{n}^{\infty}{\braces{x} \over x^{3}}\,\dd x < {1 \over n^{2}}}$ such that \eqref{1} becomes: $$\bbx{\ds{% \lim_{n \to \infty}\bracks{\pars{{\pi^{2} \over 6} - \sum_{k = 1}^{n}{1 \over k^{2}}}n} = 1}} $$

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