# If the positive series $\sum a_n$ diverges and $s_n=\sum\limits_{k\leqslant n}a_k$ then $\sum \frac{a_n}{s_n}$ diverges as well

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.

• 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 $\ln a(x)$... – vonbrand May 11 '13 at 23:16
• @vonbrand Wouldn't this be $\ln$ of the integral $\int a(t)dt$? – Eric Auld Jul 4 '13 at 20:32
• How can one do this by comparison with integrals? – JLA Jan 11 '14 at 5:52
• Stolz...$\mbox{}$ – Felix Marin Jul 29 '14 at 0:12

## 2 Answers

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.

• Awesome! Thanks for the quick and clear answer. – SwifferCat May 11 '13 at 23:06
• Thanks. By the way, do you know why the very first assertion of my answer (the existence of $m$ such that...) holds? – Did May 12 '13 at 12:12
• @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) $1/2$? – rbird Dec 27 '18 at 12:11
• @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. – Did Dec 27 '18 at 17:18

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 $$\sum\frac{a_n}{s_n}$$.