A Sum of a Double Series

Dorin Marghidanu posted the following statement and its proof at the CutTheKnotMath facebook page:

If $m,n\in\mathbb{N},$ $m,n\ge 2,$ find the best constant $k\in\mathbb{R},$ for which

$\displaystyle\sum_{j=2}^{n}\sum_{i=2}^{m}\frac{1}{i^{j}}\lt k.$

Solution

The sequence $s_{m,n}=\displaystyle\sum_{j=2}^{n}\sum_{i=2}^{m}\frac{1}{i^{j}}$ is clearly increasing in both indices $m$ and $n$ such that the best constant $k,$ if exists, is obtained by passing to the limit when $m,n\rightarrow\infty.$ The derivation includes changing the order of summation (of finite sums) and using the formula for the sum of a geometric series:

$\begin{align} \displaystyle\lim_{m,n\rightarrow\infty}s_{m,n}&=\displaystyle\lim_{m,n\rightarrow\infty}\sum_{j=2}^{n}\sum_{i=2}^{m}\frac{1}{i^{j}}\\ &=\displaystyle\lim_{m,n\rightarrow\infty}\sum_{i=2}^{m}\sum_{j=2}^{n}\frac{1}{i^{j}}\\ &=\displaystyle\lim_{m,n\rightarrow\infty}\sum_{i=2}^{m}\left(\frac{1}{i^{2}}\frac{1-\displaystyle\frac{1}{i^{n-1}}}{1-\displaystyle\frac{1}{i}}\right)\\ &=\displaystyle\lim_{m\rightarrow\infty}\sum_{i=2}^{m}\frac{1}{i(i-1)}\\ &=\displaystyle\lim_{m\rightarrow\infty}\sum_{i=2}^{m}\left(\frac{1}{i-1}-\frac{1}{i}\right)\\ &=\displaystyle\lim_{m\rightarrow\infty}\left(1-\frac{1}{m}\right)\\ &= 1. \end{align}$

Therefore, the best constant is $k=1.$

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