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$$\sum_{m=2}^{10^6}\ \sum_{n=1}^{\infty}\dfrac{m^{\{n\}}+m^{-{\{n\}}}}{m^n}$$

Where $\{n\}=\text{ integer nearest to the } \sqrt{n} $.

Looks like a non-trivial summation. How to properly approach this problem?

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    $\begingroup$ Please do typeset this in LaTeX. $\endgroup$
    – M.B.
    Commented Jan 26, 2011 at 10:36
  • $\begingroup$ Sorry but I am new to this site and Latex. So don't have much experience with it. $\endgroup$ Commented Jan 26, 2011 at 10:40
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    $\begingroup$ The inner sum doesn't converge... $m^{\{n\}}/m^n$ is asymptotic to $m^{-1/2}$, whose partial sums diverge. $\endgroup$
    – mjqxxxx
    Commented Jan 26, 2011 at 20:39

2 Answers 2

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Define $$ S \left( T \right) := \sum_{m = 2}^T \sum_{n = 1}^{\infty} \frac{m^{\left\{ n \right\}} + m^{- \left\{ n \right\}}}{m^n} . $$ First, let's find all numbers n such that $\left\{ n \right\} = t$ for some integer t: $$ \sqrt{n} - \frac{1}{2} < t < \sqrt{n} + \frac{1}{2} \Leftrightarrow t^2 - t + 1 \leq n \leq t^2 + t $$ and so, $$ S \left( T \right) = \sum_{m = 2}^T \sum_{t = 1}^{\infty} \sum_{n = t^2 - t + 1}^{t^2+t} \frac{m^t + m^{-t}}{m^n} = \sum_{m = 2}^T \sum_{t = 1}^{\infty} \frac{\left(m^t + m^{-t} \right) \left(m^{- t^2 + t - 1} - m^{-t^2 - t - 1} \right) m}{m - 1} . $$ The summand can now be rewritten to yield $$ S \left( T \right) = \sum_{m = 2}^T \frac{m}{m-1} \sum_{t = 1}^{\infty} \left( m^{- \left( t - 1 \right)^2} - m^{- \left( t + 1 \right)^2} \right), $$ which is a telescopic sum in t. Thus, $$ S \left( T \right) = \sum_{m = 2}^T \frac{m}{m-1} \left( 1 + \frac{1}{m} \right) = \sum_{m = 2}^T \left( 1 + \frac{2}{m - 1} \right) = T - 1 + 2 H(T - 1), $$ where $H \left( k \right) := \sum_{i = 1}^k \frac{1}{i}$ denotes the harmonic sum.

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Summing over all $n$ from $1$ to $100$ Mathematica produces 1.0000277854514457314*10^6.

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