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I've tried to use the limit test: $\lim_{n \rightarrow \infty} \frac{a_n}{b_n} = C$ in the series of $a_n$ and $\frac{a_n}{a_1+a_2+...+a_n}$ but it was inconclusive.

The question does not especify the sign of $a_n$ but proving it for positives would already help.

I think the sign can change as long as the partial sums never get to be zero. Like if the series has only negative numbers the result seems to stil hold.

Show that the series $\sum_{n=1}^\infty \frac{a_n}{S_n}$ diverges

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    $\begingroup$ Are we to assume all terms are positive? $\endgroup$ – alex.jordan Feb 5 '15 at 21:53
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    $\begingroup$ Random thought, but the terms remind me of the identity: $$1=\sum_{k=0}^n\frac{a_k}{(a_0+1)(a_1+1)(a_2+1)(a_3+1)(a_4+1)...(a_k+1)}+\frac{1}{(a_0+1)(a_1+1)(a_2+1)(a_3+1)(a_4+1)...(a_n+1)}$$ For all $\{a_i \}$ and integers $n\ge 0$ , so long as $a_j\ne 0$ for any integer $0\leq j \leq n$ $\endgroup$ – Ethan Feb 5 '15 at 21:53
  • $\begingroup$ Are you sure ? I've tried the stolz theorem and thus I calculated $a_{n+1} -a_n$ of the series and it didn't look telescopic $\endgroup$ – onlyme Feb 5 '15 at 21:53
  • $\begingroup$ @onlyme I was thinking of something similar. $\endgroup$ – Ethan Feb 5 '15 at 21:54
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Suppose the $a_n$ are non-negative. To prove that the series diverges to $+\infty$, it is enough to prove that for every $N\in\Bbb N$, there exists $N'>N$ such that $$\sum_{n=N}^{N'}\frac{a_n}{a_0+\cdots+a_n}\geq \frac12$$ So fix $N>0$, and notice that for every $M>N$ $$\sum_{n=N}^{M}\frac{a_n}{a_0+\cdots+a_n}\geq\sum_{n=N}^{M}\frac{a_n}{a_0+\cdots+a_{M}}=\frac{a_N+\cdots+a_{M}}{a_0+\cdots+a_{M}}=1-\frac{C}{a_0+\cdots+a_{M}}$$ where $C=a_0+\cdots+a_{N-1}$ is a constant. The second term is of the form "constant over something that tends to infinity" and so tends to $0$, and eventually we get a minoration by $\frac12$, showing that the series diverges to infinity.

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  • $\begingroup$ It is not clear to me if you showed that the series diverges or if it can have a limit greater than 1. Thank you for the answer $\endgroup$ – onlyme Feb 5 '15 at 22:31
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    $\begingroup$ It diverges to $+\infty$, because say we have started with $N_0=0$, we find $N_1>N_0$ such that the sum of terms from $N_0$ to $N_1-1$ is greater than $\frac12$, and we then find $N_2$ such that the sum of terms from $N_1$ to $N_2-1$ is greater than $\frac12$, etc... After $k$ iterations the sum will of terms from $0=N_0$ to $N_k-1$ will be greater than $\frac{k}2$. $\endgroup$ – Olivier Bégassat Feb 5 '15 at 22:35
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This problem has an interesting history from the early days of real analysis attached to it. I really like this story and the proof it contains of the problem in this question is also different from the other answers posted.


In 1827 a M. Oliver published a proof in Crelle's Journal (Journal für die reine und angewandte Mathematik) of the following theorem:

Theorem M. Oliver: A series of positive terms converges if and only if $\lim_{n\to\infty} na_n = 0$.

Not long after this paper had been published Niels Henrik Abel responded with a short note noting that the theorem "does not seem to hold true" as the series $\sum_{n=2}^\infty\frac{1}{n\log(n)}$ diverges but $\lim_{n\to\infty}n\cdot \frac{1}{n\log(n)} = 0$ contradicting the theorem above.

Abel then followed it up by another short note showing that there cannot exist any function $\phi(n)$ such that

$$\sum_{k=1}^\infty a_n ~~~\text{converges} \iff \lim_{n\to\infty}\phi(n)a_n = 0$$

The proof is remarkably simple and uses the lemma below (which coincides with the problem this question asks for). I will here present Abel's original proof. It's from memory so it's not word for word but the main idea behind the proof should be intact.


Lemma: if a series of positive terms $\sum_{k=1}^\infty a_k$ diverges then so does the series $\sum_{k=1}^\infty \frac{a_k}{a_1+a_2+\ldots + a_{k-1}}$.

Proof: Let $s_n = \sum_{k=1}^{n}a_k$, then

$$\log\left(\frac{s_k}{s_{k-1}}\right) = \log\left(1+\frac{a_k}{s_{k-1}}\right) < \frac{a_k}{s_{k-1}}$$

since $\log(1+x) < x$. Summing over $k=2,3,\ldots,n$ the left hand side telescopes to give

$$\log\left(\frac{s_n}{s_{1}}\right) < \sum_{k=1}^n \frac{a_k}{s_{k-1}}$$

Now if $s_n = \sum_{k=1}^n a_k$ diverges then so does the logarithm of $s_n$ and it follows that $\sum_{k=1}^n \frac{a_k}{s_{k-1}}$ also diverges.


Proof of Abel's theorem above: Assume there exist a function $\phi(n)$ with the properties that $\sum a_k$ converges if and only if $\lim_{n\to\infty}\phi(n)a_n = 0$. Then $\sum_{k=1}^\infty \frac{1}{\phi(k)}$ diverges since $\lim_{n\to\infty} \phi(n)\frac{1}{\phi(n)} = 1$. By the lemma above this implies that $$\sum_{k=1}^\infty \frac{\frac{1}{\phi(k)}}{\frac{1}{\phi(1)}+\frac{1}{\phi(2)} + \ldots + \frac{1}{\phi(k-1)}}$$ also diverges but $$\lim_{n\to\infty} \phi(n)\frac{\frac{1}{\phi(n)}}{\frac{1}{\phi(1)}+\ldots + \frac{1}{\phi(n-1)}} = \lim_{n\to\infty}\frac{1}{\frac{1}{\phi(1)}+\ldots + \frac{1}{\phi(n-1)}} = 0$$

which would indicate that it is convergent. No such function $\phi(n)$ can therefore exist.

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Intuition comes from the continuous analogue: if $f : [a, \infty) \to \Bbb{R}^+$ is a positive piecewise continuous function such that $\int_{a}^{\infty} f(x) \, dx = \infty$, then

$$ \int_{a}^{\infty} \frac{f(x)}{F(x)} \, dx = \log F(\infty) - \log F(a) = \infty $$

where $F$ is an antiderivative of $f$ (which is also positie on $[a, \infty)$. Bringing this idea to discrete setting is not hard. Indeed, when $a_k$'s are positive then you may apply Stolz theorem to

$$ \lim_{n\to\infty} \frac{\sum_{k=1}^{n} (a_k / s_k)}{\log s_n}. \tag{1} $$

More precisely, we have either $\limsup_{n\to\infty} a_n / s_n > 0$ or that (1) converges to 1, and in either case the series diverges.

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  • $\begingroup$ I like your answer a lot, if I can prove that $\lim_{n \rightarrow \infty} \frac{a_n}{s_n} = 0 $ or at least that it exists it's done. I've tried using the limit test again but this was useless. $\endgroup$ – onlyme Feb 6 '15 at 1:50
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    $\begingroup$ @onlyme, If $a_{n}/s_{n} \nrightarrow 0$ then the series just diverges. So you may assume $a_{n}/s_{n} \to 0$ and show that the series still diverges by examining (1). (Actually both cases can arise.) $\endgroup$ – Sangchul Lee Feb 6 '15 at 3:12
  • $\begingroup$ true, thanks a lot man. $\endgroup$ – onlyme Feb 6 '15 at 5:18
  • $\begingroup$ Anyone have anything for the case where not all the $a_n$ are non-negative? $\endgroup$ – DanielWainfleet Nov 9 '15 at 0:17
  • $\begingroup$ @user254665 If $\sum_na_n\to\infty$, then the $a_n$ are eventually nonnegative and you can use essentially the same argument as this or Oliver's answer. If $\sum_na_n\to-\infty$ by symmetry the same proof shows that $\sum_na_n/s_n\to-\infty$, so the theorem as stated is false. If $\sum_na_n$ doesn't converge to anything, the theorem is false: for example if $(a_n)=1,-3,6,-3,-3,6,-3,\dots$ (period 3 with one initial term) then $(s_n)=1,-2,4,1,-2,4,\dots$ and $(a_n/s_n)=1,3/2,3/2,-3,3/2,3/2,-3,\dots$, so $\sum_na_n/s_n$ cycles through $1,5/2,4$ without converging. $\endgroup$ – Mario Carneiro Jul 20 '16 at 7:52

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