A series with shifted index still converges Suppose that $\sum_{k=1}^{\infty} a_k$ converges.  Show that 
$\sum_{k=s}^{\infty} a_k$ for every $s > 0$  also converges
I thought about letting $x_m = \sum_{k=1}^{m} a_k$ and $y_m = \sum_{k=s}^{m}$, and showing that $x_m \to 0$ and $y_m \to 0$. But not sure what would change with the new index s.
 A: Let $s>1$,then
$$y_m=x_m-\sum_{k=1}^{s-1}a_k$$ and $x_m$ converges and $\sum_{k=1}^{s-1}a_k$ is a constant number.
A: Consider the following:
$\sum_{n=1}^{\infty} a_n = \sum_{n=1}^{s-1} a_n + \sum_{n=s}^{\infty}$
because 
$\sum_{n+1}^{\infty} a_n $
is already known to converge, we will write:
$c_1= \sum_{n=1}^{s-1} a_n + \sum_{n=s}^{\infty}$
Consider that, if a single $a_n$ present in the sum is undefined (or $lim_{n->\infty} a_n = \infty$, then $\sum_{n=1}^{\infty} a_n$ would diverge. We know, however, that this series doesn't diverge; hence, by contradiction $a_n$ does not contain any undefined values
then, since $a_n$ must be a series of finite constants, and 
$ \sum_{n=1}^{s-1} a_n$
 is a sum of finitely many terms, this series is a finite sum of constants; hence, another constant. We will denote this constant as $c_2$, then
$c_1 = c_2+ \sum_{n=s}^{\infty} a_n$ or
$c_1-c_2 = \sum_{n=s}^{\infty} a_n$.
The series is, thusly, a difference of constants; hence, the series is a constant; hence, the series converges.
