Formula for $\frac{1}{(n)^2}+\frac{1}{(n-1)^2}+\dots+\frac{1}{1^2}$.

Question

I have this question when I try to solve some problem.

We know $\frac{1}{(n)}+\frac{1}{(n-1)}+\dots+\frac{1}{1}\to(\log n)+\lambda$, as $n\to \infty$, where $\lambda$ is a constant. What is the value of $\frac{1}{(n)^2}+\frac{1}{(n-1)^2}+\dots+\frac{1}{1^2}$?

More generally, what's $ \sum_{m=1}^{n-1}\frac{1} {m^r} \frac{1} {(n-m)^{s}}$, as $n\to\infty$, where r+s is a positive natural number? (or a simpler question, when r=s.) Is there a formula for it? (Or a formula for its lower bound more accurate than $(n-1)(\frac{2}{n})^{2\max(r,s)}$ (obtained by using geometric mean being smaller than arithmetic mean?) Furthermore, what’s the case when r+s is a constant integer, or r-s is a constant integer?

Answer 1

Well, this is the Basel Problem. We have $$ \lim_{n \to \infty} \Bigg[ \sum_{k = 1}^{n} \frac{1}{k^2} \Bigg] = \frac{\pi^2}{6} $$

And what you mentioned in your post, $$ \lim_{n \to \infty} \Bigg[ \sum_{k = 1}^n \frac{1}{k} - \log(n) \Bigg] = \gamma $$ Where $\gamma$ is the Euler-Mascheroni Constant.

Answer 2

$$S'=\sum_{k=2}^{n} \frac{1}{k^2} < \sum_{k=2}^n \frac{1}{k(k-1)}= \sum_{k=2}^{n}\left(\frac{1}{k-1}-\frac{1}{k}\right)$$ By telescopic summation, we get $$S'<1-\frac{1}{n}.$$ Hence, $$\sum_{k=1}^{n} \frac{1}{k^2}< 2-\frac{1}{n}$$ The sum of yhe infinite version of this series is well known as $\frac{\pi^2}{6}$.

Answer 3

To answer your last question, we have that

$$(n-1)^{1-\max(r,s)} \leq \sum_{m=1}^{n-1}\frac{1}{m^r}\frac{1}{(n-m)^s} \leq (n-1)^{1-\min(r,s)}$$

WLOG, we can assume $s \leq r$ (since the sum is symmetric when summing backwards). So we see that the sum only exists when $s\geq 1$

Answer 4

There is an asymptotic expansion in terms of the Bernoulli numbers: $$ \sum\limits_{k = 1}^n {\frac{1}{{k^2 }}} \sim \frac{{\pi ^2 }}{6} - \frac{1}{n} + \frac{1}{{2n^2 }} - \sum\limits_{k = 1}^\infty {\frac{{B_{2k} }}{{n^{2k + 1} }}} $$ as $n\to +\infty$. This shows how the limit $\frac{\pi^2}{6}$ is approached.

Answer 5

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $\ds{\LARGE\left.a\right)}$

With the Abel-Plana Formula and $\ds{\vartheta_{n - 1} \in \pars{0,1}}$: \begin{align} \sum_{k = 1}^{n}{1 \over k^{2}} & = \sum_{k = 0}^{n - 1}{1 \over \pars{k + 1}^{2}} \\[5mm] & = \int_{0}^{n - 1}{\dd x \over \pars{x + 1}^{2}} + \bracks{{1 \over 2}\,{1 \over \pars{k + 1}^{2}}}_{\ k\ =\ 0} + \bracks{{1 \over 2}\,{1 \over \pars{k + 1}^{2}}}_{\ k\ =\ n - 1} \\[2mm] & +\ \underbrace{4\int_{0}^{\infty}{x \over \pars{x^{2} + 1}^{2}\pars{\expo{2\pi x} - 1}}\,\dd x} _{\ds{{\pi^{2} \over 6} - {3 \over 2}}}\ +\ \sum_{s = 1}^{m}{B_{2s} \over \pars{2s}!} \bracks{-\,{\pars{2s}! \over n ^{2s + 1}}} \\[2mm] & + 2\pars{-1}^{m}\pars{2m + 1}\int_{0}^{\infty} \Im\pars{\bracks{n + \ic\vartheta_{n - 1}x}^{-2m - 2}}\, {x^{2m} \over \expo{2\pi x} - 1}\,\dd x \\[5mm] & = \pars{1 - {1 \over n}} + {1 \over 2} + {1 \over 2n^{2}} + \pars{{\pi^{2} \over 6} - {3 \over 2}} - \sum_{s = 1}^{m}{B_{2s} \over n^{2s + 1}} \\[2mm] & + 2\pars{-1}^{m}\pars{2m + 1}\int_{0}^{\infty} \Im\pars{\bracks{n + \ic\vartheta_{n - 1}x}^{-2m - 2}}\, {x^{2m} \over \expo{2\pi x} - 1}\,\dd x \\[5mm] & = \bbox[10px,#ffd]{{\pi^{2} \over 6} - {1 \over n} + {1 \over 2n^{2}} + \sum_{s = 1}^{m}{B_{2s} \over n^{2s + 1}}} \\[2mm] & \bbox[10px,#ffd]{+\ 2\pars{-1}^{m}\pars{2m + 1}} \\[1mm] & \bbox[10px,#ffd]{\left. \phantom{=}\times\int_{0}^{\infty} \Im\pars{\bracks{n + \ic\vartheta_{n - 1}x}^{-2m - 2}} {x^{2m} \over \expo{2\pi x} - 1} \,\dd x\,\right\vert_{\, \vartheta_{n - 1}\ \in\ \pars{0,1}}} \end{align}

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$\ds{\LARGE\left.b\right)}$

By using a Zeta function identity: \begin{align} \sum_{k = 1}^{n}{1 \over k^{\color{red}{2}}} & = \zeta\pars{2} - \,{n^{1 - \color{red}{2}} \over \color{red}{2} - 1} + \color{red}{2}\int_{n}^{\infty} {x - \left\lfloor\,{x}\,\right\rfloor \over x^{\color{red}{2} + 1}} \,\dd x \\[5mm] & = {\pi^{2} \over 6} - {1 \over n} + 2\sum_{k = n}^{\infty}\int_{k}^{k +1}{x - k \over x^{3}}\,\dd x \\[5mm] & = {\pi^{2} \over 6} - {1 \over n} +\ \underbrace{\sum_{k = n}^{\infty}{1 \over k\pars{k + 1}^{2}}} _{\ds{1 + n - n^{2}\,\Psi\, '\pars{n} \over n^{2}}}\quad \pars{~\Psi\, ':\ Trigamma\ Function~} \\[5mm] & = \bbox[10px,#ffd]{{\pi^{2} \over 6} - {1 \over n} + {1 \over 2n^{2}} - {1 \over 6n^{3}} + {1 \over 30n^{5}} + \mrm{O}\pars{1 \over n^{6}}} \end{align}