Some Mistakes can give rise to problem

$\large{\displaystyle \sum^{\infty}_{n=1}\frac{2n}{(2n-1)e^{(2n-1)}}}$

$\large{=\displaystyle \sum^{\infty}_{n=1} \left( \frac{1}{e^{2n-1}}+\frac{1}{(2n-1)e^{2n-1}} \right)}$

$\large{\frac{\frac{1}{e}}{1-\frac{1}{e^{2}}}+\frac{1}{e}+\frac{1}{3e^3}+\frac{1}{5e^5}+....}$

we know

$\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-......$

$\large{\frac{1}{2}\ln \left(\frac{1+x}{1-x} \right)=x+\frac{x^3}{3}+\frac{x^5}{5}+.....}$

The summation becomes

$\large{\frac{1}{2}\ln \left(\frac{e+1}{e-1} \right)+\frac{e}{e^2-1}=\boxed{0.811}}$

Bonus question- Try yourself

$Take\quad an\quad infinite\quad GP\quad \frac { 1 }{ x } ,\frac { 1 }{ { x }^{ 3 } } ,\frac { 1 }{ { x }^{ 5 } } ,\frac { 1 }{ { x }^{ 7 } } ...\quad |x|>1\\ We\quad can\quad easily\quad see\quad that\\ \quad \quad \quad \quad \frac { x }{ { x }^{ 2 }-1 } \quad =\quad \frac { 1 }{ x } +\frac { 1 }{ { x }^{ 3 } } +\frac { 1 }{ { x }^{ 5 } } +...\\ \Rightarrow \quad \quad \quad \frac { 1 }{ { x }^{ 2 }-1 } =\quad \frac { 1 }{ { x }^{ 2 } } +\frac { 1 }{ { x }^{ 4 } } +\frac { 1 }{ { x }^{ 6 } } +...\\ \quad \int _{ 0 }^{ e }{ \frac { 1 }{ { x }^{ 2 }-1 } dx= } \int _{ 0 }^{ e }{ \left( \frac { 1 }{ { x }^{ 2 } } +\frac { 1 }{ { x }^{ 4 } } +\frac { 1 }{ { x }^{ 6 } } +... \right) dx } \\ \Rightarrow \frac { 1 }{ 2 } \ln { \left( \frac { e+1 }{ e-1 } \right) } =\quad \sum _{ n=1 }^{ \infty }{ \frac { 1 }{ (2n-1){ e }^{ 2n-1 } } }$