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say we have a sequence of non-negative reals, $a_1, a_2, \dots$, and that $\displaystyle\sum\limits_{n=1}^{\infty}a_n$ is divergent, meaning convergent to infinity. Under this scenario I am trying to prove that the following sequence in $m$ cannot converge to zero.

$$t_m \,\,=\,\, \displaystyle\sum\limits_{n=1}^{m} \frac{n}{m}a_n$$

I'd like to know if this proposition is true. I was hoping so, but became stuck trying to prove it. My reasoning so far:

Since $\Sigma a_n \,=\, +\infty$, the sequence of partial sums is not Cauchy. So there exists an $\epsilon$ and indices $i>j>0$ for which $$a_{j+1} + a_{j+2} + \dots + a_{i} \,\, \geq \epsilon.$$

But then we can say there is an infinite sequence of such finite segments; we can always produce another one. Now look at the sequence $t$, e.g.

$$t_5 \,\,=\,\, \frac{1}{5}a_1 \,+\,\frac{2}{5}a_2 \,+\,\frac{3}{5}a_3 \,+\,\frac{4}{5}a_4 \,+\,\frac{5}{5}a_5 \,\,\geq\,\,\frac{1}{2}a_3 \,+\,\frac{1}{2}a_4 \,+\,\frac{1}{2}a_5.$$ Thus in general:

$$t_m \,\,\geq\,\,\frac{1}{2}\displaystyle\sum\limits_{[m/2]+1}^{m}a_n.$$

Is there any hope to tie this to the sequence of epsilon segments above, and show that my sequence $t$ is strictly away from zero? It seems a little reasonable, since as $m$ grows big, $t_m$ is a sum of many many terms, arbitrarily many. It would suffice to show that infinitely often, my $t_m$ is at least the fixed positive epsilon.

311411
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1 Answers1

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A counterexample is given by $a_n=(n\log n)^{-1}$ for $n>1$ (and, say, $a_1=0$). The series $\sum_{n=1}^{\infty}a_n$ is well-known to diverge, while $t_m=m^{-1}\sum_{n=1}^{m}(\log n)^{-1}\underset{m\to\infty}{\longrightarrow}0$ by the Stolz–Cesàro theorem.

metamorphy
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  • so we may use the $\frac{.}{\infty}$ case of the theorem, setting $B_m = m$ and $A_m=\Sigma{\frac{1}{\log n}}$. Then $\lim \frac{A_m}{B_m} = \lim t_m = 0$ holds if $\lim \frac{1}{\log (m+1)} = 0$ holds. That your example series diverges we can see from the Integral Test, using $F(x) = \log (\log x)$. Thanks! – 311411 Jun 20 '20 at 15:21
  • Yes, exactly. In fact any $a_n$ with $na_n\to 0$ and a divergent $\sum a_n$ would do, for the same reason. – metamorphy Jun 20 '20 at 15:22