# 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 $j>1$.

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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.

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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.

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It is easy to show that: $$\sum_{k=j}^{\infty}a_k=\sum_{k=1}^{\infty}a_k-(a_1+a_2+...+a_{j-1})$$