# convergent series where not only$~{}a_{n}$ but also $na_{n}$ tends to 0

Proposition.  If the terms (http://planetmath.org/Series) $a_{n}$ of the convergent series

 $a_{1}+a_{2}+\ldots$

are positive and form a monotonically decreasing sequence, then

 $\displaystyle\lim_{n\to\infty}na_{n}\;=\;0.$ (1)

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

 $0\;<\;a_{m+1}+\ldots+a_{m+p}\;<\;\frac{\varepsilon}{2}\qquad(p\;=\;1,\,2,\,% \ldots).$

Since the sequence  $a_{1},\,a_{2},\,\ldots$  is decreasing, this implies

 $\displaystyle 0\;<\;pa_{m+p}\;<\;\frac{\varepsilon}{2}\qquad(p\;=\;1,\,2,\,% \ldots).$ (2)

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

 $0\;<\;ma_{m+m}\;<\;\frac{\varepsilon}{2},$

whence again due to the decrease,

 $\displaystyle 0\;<\;ma_{m+p}\;<\;\frac{\varepsilon}{2}\qquad(p\;=\;m,\,m\!+\!1% ,\,\ldots).$ (3)

Adding the inequalities (2) and (3) with the common values  $p=m,\,m\!+\!1,\,\ldots$  then yields

 $0\;<\;(m\!+\!p)a_{m+p}\;<\;\varepsilon\qquad\mbox{for}\quad p\;\geqq\;m.$

This may be written also in the form

 $0\;<\;na_{n}\;<\;\varepsilon\qquad\mbox{for}\quad n\;\geqq\;2m$

which means that  $\lim_{n\to\infty}na_{n}\;=\;0$.

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 $a_{1}\!+\!a_{2}\!+\ldots$ 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  $na_{n}=1$  for each perfect square $n$; so  $na_{n}\not\to 0$  as  $n\to\infty$.

Title convergent series where not only$~{}a_{n}$ but also $na_{n}$ tends to 0 ConvergentSeriesWhereNotOnlyanButAlsoNanTendsTo0 2013-03-22 19:03:29 2013-03-22 19:03:29 pahio (2872) pahio (2872) 14 pahio (2872) Theorem msc 40A05 Olivier’s theorem NecessaryConditionOfConvergence AGeneralisationOfOlivierCriterion