if $\sum \mathrm{@}\mathrm{@}\mathrm{@}{\mathrm{@}}_{k=1}^{\mathrm{\infty }}{a}_k$ converges then $a_k\to 0$

Theorem 1.

Suppose $a_{\mathrm{1}}\mathrm{,}{a}_{\mathrm{2}}\mathrm{,}\mathrm{\dots }$ is a sequence of real or complex numbers.If the series

$$\sum _{k=1}^{\mathrm{\infty }}{a}_k$$

converges, then ${\lim }_{k\mathrm{\to }\mathrm{\infty }}\mathit{}{a}_k\mathrm{=}\mathrm{0}$.

Remarks

1. The harmonic series ${\sum }_{k=1}^{\mathrm{\infty }}1/k$ shows that theimplication can not be reversed.

2. This result can be used as a first test for convergence of a series${\sum }_{k=1}^{\mathrm{\infty }}{a}_k$. If $a_k$ does not converge to $0$, then${\sum }_{k=1}^{\mathrm{\infty }}{a}_k$ does not converge either.

Proof.

Let $S\in ℂ$ be the value of the sum, and let $\epsilon >0$be arbitrary. Then there exists an $N\ge 1$ such that

for all $M\ge N$. For $j\ge N$ we then have

$|a_{j+1}|$$=$$|{\displaystyle \sum _{k=1}^{j+1}}{a}_k-{\displaystyle \sum _{k=1}^j}{a}_k|$

$\le$$|{\displaystyle \sum _{k=1}^{j+1}}{a}_k-S|+|S-{\displaystyle \sum _{k=1}^j}{a}_k|$

$\epsilon ,$

and the claim follows.∎