Q:

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:

$$\sum _{ k=2 }^{ \infty  } \sum _{ n=2 }^{ \infty  } \frac { 1 }{ k^{ n } } =\sum _{ k=2 }^{ \infty  } \left( \frac { 1 }{ { k }^{ 2 } } +\frac { 1 }{ { k }^{ 3 } } +... \right) =\sum _{ k=2 }^{ \infty  } \frac { \frac { 1 }{ { k }^{ 2 } }  }{ 1-\frac { 1 }{ k }  } =\sum _{ k=2 }^{ \infty  } \frac { 1 }{ k\left( k-1 \right)  } =\\ =\sum _{ k=2 }^{ \infty  } \left( \frac { 1 }{ k-1 } -\frac { 1 }{ k }  \right) =1$$

