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So I've been trying to figure out how to prove the following.

Let $(a_n)$ be a sequence of positive numbers such that $\sum\limits_{n=1}^\infty a_n =\infty$, and define $s_n=\sum\limits_{i=1}^n a_i$. Then $\sum\limits_{n=1}^\infty\frac{a_n}{s_n} =\infty$ as well.

I can prove it by comparing it to $\int_1^\infty \frac{1}{x} \, dx $ if the sequence $a_n$ is bounded by some $M$, but that's as far as I've been able to get.

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    $\begingroup$ Your $\sum \frac{a_n}{s_n}$ is an awful lot like $\int \dfrac{a(x) dx}{\int a(t) dt}$, i.e. it smells of $\ln a(x)$... $\endgroup$
    – vonbrand
    Commented May 11, 2013 at 23:16
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    $\begingroup$ @vonbrand Wouldn't this be $\ln$ of the integral $\int a(t)dt$? $\endgroup$
    – Eric Auld
    Commented Jul 4, 2013 at 20:32
  • $\begingroup$ How can one do this by comparison with integrals? $\endgroup$
    – JLA
    Commented Jan 11, 2014 at 5:52
  • $\begingroup$ Stolz...$\mbox{}$ $\endgroup$ Commented Jul 29, 2014 at 0:12

2 Answers 2

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Since $s_n\to+\infty$ when $n\to\infty$, for each $n\geqslant1$ there exists some finite $m\gt n$ such that $s_m\geqslant2s_n$. In particular, $\sum\limits_{k=n+1}^m\frac{a_k}{s_k}\geqslant\sum\limits_{k=n+1}^m\frac{a_k}{s_m}=1-\frac{s_n}{s_m}\geqslant1/2$. Thus, the rests of the series $\sum\limits_k\frac{a_k}{s_k}$ do not converge to zero, QED.

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  • $\begingroup$ Awesome! Thanks for the quick and clear answer. $\endgroup$
    – SwifferCat
    Commented May 11, 2013 at 23:06
  • $\begingroup$ Thanks. By the way, do you know why the very first assertion of my answer (the existence of $m$ such that...) holds? $\endgroup$
    – Did
    Commented May 12, 2013 at 12:12
  • $\begingroup$ @Did What does the last bit mean? Why does it help to show $\sum\frac{a_k}{s_k}$ do not converge to zero? Could they not converge to (say) $1/2$? $\endgroup$
    – A. Goodier
    Commented Dec 27, 2018 at 12:11
  • $\begingroup$ @rbird Please quote accurately: the rests of the series do not converge to zero, which is a well-known (necessary and) sufficient condition of divergence of a series. $\endgroup$
    – Did
    Commented Dec 27, 2018 at 17:18
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Alternative proof 1, using the inequality $x\ge \log(x+1)$: $$ \frac{a_n}{s_n}\ge\log\left(1+\frac{a_n}{s_n}\right)=\log\left(\frac{s_n+a_n}{s_n}\right)=\log(s_{n+1})-\log(s_n)\tag1 $$ Summing (1) from $n=1$ to $n=N$, the RHS telescopes: $$ \sum_{n=1}^N\frac{a_n}{s_n}\ge\log(s_{N+1})-\log(s_1). $$ But $s_n\to\infty$, so the series $\sum\frac{a_n}{s_n}$diverges.


Alternative proof 2, using theory of integration (apologies for the heavy machinery): Define the sequence $(f_n)$ of functions on $[0,\infty)$ by $f_n:=\frac1{s_n} I_{[0,s_n]}$. (So $f_n$ is a rectangle of height $\frac1{s_n}$ placed over the interval $[0,s_n]$.) Then $\int f_n=1$ for each $n$. But $f_n\to0$ pointwise (since $s_n\to\infty$), so $\int\lim f_n=0$. By the dominated convergence theorem, any $g$ that satisfies $|f_n|\le g$ for all $n$ must have $\int g=\infty$. Apply this conclusion to $g:=\sum_{n=1}^\infty\frac1{s_n}I_{[s_{n-1},s_n]}$, which has integral $\sum\frac{a_n}{s_n}$.

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    $\begingroup$ $s_n+a_n\neq s_{n+1}$. $\endgroup$
    – Riemann
    Commented Apr 30, 2021 at 10:02
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    $\begingroup$ You can only get $$\frac{a_n}{s_{n-1}}\ge\log\left(1+\frac{a_n}{s_{n-1}}\right)=\log\left(\frac{s_{n-1}+a_n}{s_{n-1}}\right)=\log s_n-\log s_{n-1}.$$ $\endgroup$
    – Riemann
    Commented Apr 30, 2021 at 10:37

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