Suppose that $\Sigma_{n=1}^\infty a_n$ converges. Then $\Sigma_{n=k}^\infty a_n$ also converges for any $k>1$. So my question is, in such a case, does the limit of $\Sigma_{n=k}^\infty a_n$ as $k$ goes to $\infty$ always equal $0$?
I think the answer is yes, because $\Sigma_{n=k}^\infty a_n=\Sigma_{n=1}^\infty a_n-\Sigma_{n=1}^{k-1} a_n$, and the right hand side goes to $\Sigma_{n=1}^\infty a_n-\Sigma_{n=1}^\infty a_n=0$ as $k$ goes to $\infty$. But I just wanted to double check.
Yes it does.
Note that if $$ \Sigma_{n=1}^\infty =S$$ then $$\Sigma_{n=k}^\infty a_n=\Sigma_{n=1}^\infty a_n-\Sigma_{n=1}^{k-1} a_n =S-\Sigma_{n=1}^{k-1} a_n$$
Thus if $k\to \ $$\infty$ we get $$ \lim_{k\to \infty}\Sigma_{n=k}^\infty a_n =S-S=0$$