Get the solution to "Summation of a non trivial series∑m=2106∑n=1∞mn+m−nmnwhere n=integer nearest..."

Pablo Dennis

Answered question

2022-04-02

Summation of a non trivial series

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

where

n = integer nearest to the \sqrt{n}

Ashley Olson

Beginner2022-04-03Added 12 answers

Define

S (T) = \sum_{m = 2}^{T} \sum_{n = 1}^{\infty} \frac{m^{n} + m^{- n}}{m^{n}}

First, let's find all numbers n such that

[n] = 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 (T) = \sum_{m = 2}^{T} \sum_{t = 1}^{\infty} \sum_{n = t^{2} - t + 1}^{t^{2} + 1} \frac{m^{t} + m^{- t}}{m^{n}}

= \sum_{m = 2}^{T} \sum_{t = 1}^{\infty} \frac{(m^{t} + m^{- t}) (m^{- t^{2} + t - 1} - m^{- t^{2} - t + 1}) m}{m - 1}

The summand can now be rewritten to yield

S (T) = \sum_{m = 2}^{T} \frac{m}{m - 1} \sum_{t = 1}^{\infty} (m^{- {(t - 1)}^{2}} - m^{- {(t + 1)}^{2}})

which is a telescopic sum in t. Thus,

S (T) = \sum_{m = 2}^{T} \frac{m}{m - 1} (1 + \frac{1}{m}) = \sum_{m = 2}^{T} (1 + \frac{2}{m - 1}) = T - 1 + 2 H (t - 1)

where

H (k) = \sum_{i = 1}^{k} \frac{1}{i}

denotes the harmonic sum.

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