convergent series where not only$a_n$ but also $n{a}_n$ tends to 0

Proposition.  If the terms (http://planetmath.org/Series) $a_n$ of the convergent series

$$a_1+a_2+\mathrm{\dots }$$

are positive and form a monotonically decreasing sequence, then

$\underset{n\to \mathrm{\infty }}{lim}n{a}_n=\mathrm{\hspace{0.33em}0}.$(1)

Proof.  Let $\epsilon$ be any positive number.  By the Cauchy criterion for convergence and the positivity of the terms, there is a positive integer $m$ such that

Since the sequence  $a_1,a_2,\mathrm{\dots }$  is decreasing, this implies

(2)

Choosing here especially  $p:=m$,  we get

whence again due to the decrease,

(3)

Adding the inequalities (2) and (3) with the common values  $p=m,m+1,\mathrm{\dots }$  then yields

This may be written also in the form

which means that  ${lim}_{n\to \mathrm{\infty }}n{a}_n=\mathrm{\hspace{0.33em}0}$.

Remark.  The assumption of monotonicity in the Proposition is essential.  I.e., without it, one cannot gererally get the limit result (1).  A counterexample would be the series $a_1+a_2+\mathrm{\dots }$ where $a_n:=\frac{1}{n}$  for any perfect square $n$ but 0 for other values of $n$.  Then this series is convergent (cf. the over-harmonic series), but  $n{a}_n=1$  for each perfect square $n$; so  $n{a}_n\to ̸0$  as  $n\to \mathrm{\infty }$.