Evaluate $\sum_{k=2}^{\infty}\sum_{n=2}^{\infty}\frac{1}{k^n}$ How do I evaluate the $n$-sum? I know how to do it if the sum started with $n=0$ but what is problematic is the $n=2$.
$$\sum_{k=2}^{\infty}\sum_{n=2}^{\infty}\frac{1}{k^n}=?$$
 A: $$\sum _{ k=2 }^{ \infty  } \sum _{ n=2 }^{ \infty  } \frac { 1 }{ k^{ n } } =\sum _{ k=2 }^{ \infty  } \left( \frac { 1 }{ { k }^{ 2 } } +\frac { 1 }{ { k }^{ 3 } } +... \right) =\sum _{ k=2 }^{ \infty  } \frac { \frac { 1 }{ { k }^{ 2 } }  }{ 1-\frac { 1 }{ k }  } =\sum _{ k=2 }^{ \infty  } \frac { 1 }{ k\left( k-1 \right)  } =\\ =\sum _{ k=2 }^{ \infty  } \left( \frac { 1 }{ k-1 } -\frac { 1 }{ k }  \right) =1$$
A: Consider the series
\begin{align}
\sum_{n=1}^{\infty} \frac{t^{n}}{\Gamma(n+1)} &= e^{t} -1 \\
\sum_{k=2}^{\infty} e^{-k t} &= \frac{e^{-2t}}{1 - e^{-t}}
\end{align}
and
$$\int_{0}^{\infty} e^{-s t} \, t^{n} \, dt = \frac{\Gamma(n+1)}{s^{n+1}}$$
in such a way that:
\begin{align}
\sum_{k=2}^{\infty} \, \sum_{n=2}^{\infty} \frac{1}{k^{n}} &= \sum_{n=1}^{\infty} \, \frac{1}{\Gamma(n+1)} \, \sum_{k=2}^{\infty} \frac{\Gamma(n+1)}{k^{n+1}} \\
&= \sum_{n=1}^{\infty} \frac{1}{n!} \, \int_{0}^{\infty} t^{n} \, \frac{e^{-2t}}{1 - e^{-t}} \, dt \\
&= \int_{0}^{\infty} \frac{e^{-2t}}{1 - e^{-t}} \, (e^{t} -1) \, dt \\
&= \int_{0}^{\infty} e^{-t} \, dt \\
&= 1.
\end{align}
A: Observe that $\sum_{n=2}^{\infty} \frac{1}{k^n} = \frac{1}{1-\frac{1}{k}}-\frac{1}{k}-1=\frac{k-1}{k}-\frac{1}{k}-1= \frac{1}{k-1}-\frac{1}{k}$.
Then, $$\sum_{k=2}^{\infty}\left(\frac{1}{k-1}-\frac{1}{k}\right) =\frac{1}{2-1}-\lim_{b\rightarrow \infty}\frac{1}{b} =1.$$
