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I need to find the geometric series of the sum described below. It is trivial to show that it converges but I'm not sure how to find the series from the equation given.
Could someone show me a technique to do so? The geometric sequence is given in the book as noted in the picture. I omitted the work showing it converges since it is easy to find.
$$\sum_{n=1}^{\infty}\left(\frac{1}{e^n}+\frac{1}{n(n+1)} \right)$$
it is a sum of "geometric" and "telescoping" serias $$\sum _{ n=1 }^{ \infty } \left( \frac { 1 }{ e^{ n } } +\frac { 1 }{ n(n+1) } \right) =\sum _{ n=1 }^{ \infty }{ \frac { 1 }{ e^{ n } } } +\sum _{ n=1 }^{ \infty }{ \left( \frac { 1 }{ n } -\frac { 1 }{ n+1 } \right) } =\frac { 1 }{ { e } } +\frac { 1 }{ { e }^{ 2 } } +...+\left( 1-\frac { 1 }{ 2 } +\frac { 1 }{ 2 } -\frac { 1 }{ 3 } +... \right) =\frac { \frac { 1 }{ e } }{ 1-\frac { 1 }{ e } } +1=\\ =\frac { 1 }{ e-1 } +1=\frac { e }{ e-1 } $$
Hints: $$ \sum_{n=1}^\infty\biggl(\frac1{e^n}+\frac1{n(n+1)}\biggr)=\sum_{n=1}^\infty\frac1{e^n}+\sum_{n=1}^\infty\frac1{n(n+1)} $$ and $$ \sum_{n=1}^N\frac1{n(n+1)}=\sum_{n=1}^N\biggl[\frac{1}{n}-\frac{1}{n+1}\biggr]=1-\frac1{N+1}. $$
HINT: use that $$\sum_{n=1}^mx^n=\frac{x \left(x^m-1\right)}{x-1}$$ and calculate $$\lim_{m\to \infty}\sum_{n=1}^mx^n$$ or prove by induction that $$\sum_{n=1}^m\frac{1}{e^n}+\frac{1}{n(n+1)}=\frac{e^{-m} \left(m e^{m+1}+e^m-m-1\right)}{(e-1) (m+1)}$$
