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

Proposition. If the terms $a_{n}$ 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 $n\to\infty$.

## Mathematics Subject Classification

40A05*no label found*

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## Comments

## A_n pro a_n

Hi experts, how could I get "a_n" instead of "A_n" in the title of

http://planetmath.org/encyclopedia/ConvergentSeriesWhereNotOnlyA_nButAls... ?

I have made many trials =o(

Jussi

## Re: A_n pro a_n

I believe I just fixed it. But I have no idea why it works. Looks like a bug to me.

Instead of "only $a_n$" I wrote "only$~a_n$".

## Re: A_n pro a_n

Hey Roger,

You have solved the problem!! Thank you very much!

The solution is quite unconventional -- especially I wonder the omitted space in "only$~a_n$".

Best regards,

Jussi