If an infinite series, $\sum_{k=1}^{\infty}a_{k}$ is convergent. Is $\sum_{k=j}^{\infty}a_{k}$ finite? If an infinite series $\sum_{k=1}^{\infty}a_{k}$ is convergent. Is $\sum_{k=j}^{\infty}a_{k}$ finite? Where $j>1$.
 A: Yes, because
$$
\sum_{k=j}^\infty a_k = \sum_{k=1}^\infty a_k - \sum_{k=1}^{j-1}a_k.
$$
The first term on the right-hand side is finite by assumption, the second because it is the sum of finitely many finite numbers.
A: Given $\displaystyle \sum_{k=1}^{\infty}a_{k}\;\;$ is convergent, then YES: for $\;j \gt 1$, $\quad(a)\quad\sum_{k=j}^\infty a_k\;\;$is finite. 
Note that the series given by $(a)$ can be expressed as the difference of sums as shown below:
$$\sum_{k=j}^\infty a_k \quad=\quad \sum_{k=1}^\infty a_k \;\;- \;\;\sum_{k=1}^{j-1}a_k.\tag{1}$$
The first sum on the right-hand side of $(1)$ is finite since we are given that this is convergent, and so finite by definition. The second sum on the right hand side of $(1)$ is finite since it's the finite sum of finitely many terms.
We conclude that $\;\displaystyle \sum_{k=j}^\infty a_k\;$ is therefore finite.
A: It is easy to show that:
$$\sum_{k=j}^{\infty}a_k=\sum_{k=1}^{\infty}a_k-(a_1+a_2+...+a_{j-1})$$
Thus, the answer is yes.
(I hope I did not interpret your question in a trivial way.)
