Summation of a non trivial series $$\sum_{m=2}^{10^6}\ \sum_{n=1}^{\infty}\dfrac{m^{\{n\}}+m^{-{\{n\}}}}{m^n}$$
Where $\{n\}=\text{ integer nearest to the } \sqrt{n} $.
Looks like a non-trivial summation. How to properly approach this problem?
 A: 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.
A: Summing over all $n$ from $1$ to $100$ Mathematica produces
1.0000277854514457314*10^6.
