# How do you prove that the limit n to oo sum e^(-n/10) cos(n/10), for n from 0 to n, is 10.5053, nearly.?

Dec 3, 2016

See below.

#### Explanation:

${e}^{- \frac{k}{10}} \left(\cos \left(- \frac{k}{10}\right) + i \sin \left(- \frac{k}{10}\right)\right) = {e}^{- \frac{k}{10}} {e}^{- i \frac{k}{10}} = {e}^{- \left(\frac{i + 1}{10}\right) k}$

so

${\sum}_{k = 0}^{\infty} {e}^{- \frac{k}{10}} \cos \left(\frac{k}{10}\right) = {R}_{e} \left({\sum}_{k = 0}^{\infty} {e}^{- \left(\frac{i + 1}{10}\right) k}\right)$

but $\left\mid {e}^{- \left(\frac{i + 1}{10}\right) k} \right\mid = {e}^{- \frac{k}{10}} < 1$ so

${\sum}_{k = 0}^{\infty} {e}^{- \left(\frac{i + 1}{10}\right) k} = \frac{0 - 1}{{e}^{- \frac{i + 1}{10}} - 1}$ so

$\frac{1}{1 - \left({e}^{- \frac{i + 1}{10}}\right)} = \frac{1}{2 - \frac{2 \sinh \left[\frac{1}{10}\right]}{{e}^{\frac{1}{10}} - C o s \left(\frac{1}{10}\right)}} - i \left(\frac{{e}^{\frac{1}{10}} S \in \left(\frac{1}{10}\right)}{1 + {e}^{\frac{1}{5}} - 2 {e}^{\frac{1}{10}} C o s \left(\frac{1}{10}\right)}\right)$

and

${R}_{e} \left(\frac{1}{1 - \left({e}^{- \frac{i + 1}{10}}\right)}\right) = \frac{1}{2 - \frac{2 \sinh \left[\frac{1}{10}\right]}{{e}^{\frac{1}{10}} - C o s \left(\frac{1}{10}\right)}} = 5.508336109787682$