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While preparing my tutoring group on (currently) measure theory and Lebesgue integration on $\mathbb R^n$, I by chance came across the following result (at least I hope this is a result and my proof is valid) which I don't recall seeing up until today.

Let $(a_n)_{n\in\mathbb N}$ be a monotonically increasing sequence of positive numbers such that $\lim_{n\to\infty}a_n=\infty$. Then $$\sum_{n=2}^\infty \frac{a_n-a_{n-1}}{a_n}=\infty\,.$$

The proof I found for this relies on dominated convergence and goes as follows:

Proof. Given the sequence $(a_n)_{n\in\mathbb N}$ in question, define a sequence of functions $(f_n)_{n\in\mathbb N}$ via $$ f_n:[0,\infty)\to [0,\infty)\qquad x\mapsto \frac1{a_n}\mathbb 1_{[0,a_n]}(x) $$ with $\mathbb 1_{A}$ being the usual indicator function for arbitrary $A\subseteq\mathbb R$. Obviously, $\int_0^\infty f_n(x)\,dx=1$ for all $n\in\mathbb N$ and $(f_n)_{n\in\mathbb N}$ converges pointwise to zero as $\lim_{n\to\infty}a_n=\infty$ by assumption. This however implies $$ \lim_{n\to\infty}\int_0^\infty f_n(x)\,dx=1\neq 0=\int_0^\infty0\,dx=\int_0^\infty\lim_{n\to\infty} f_n(x)\,dx $$ so by (the converse of) Lebesgue's dominated convergence theorem, no dominating integrable function $g$ for $(f_n)_{n\in\mathbb N}$ can exist. This in turn means that every dominating function $g$ we find has to be non-integrable. The obvious choice here is (written in a sloppy way but it should be clear how $g$ operates) $$ g:[0,\infty)\to[0,\infty)\qquad x\mapsto \begin{cases} \frac1{a_1}&\text{ if }x\in[0,a_1]\\\frac1{a_2}&\text{ if }x\in(a_1,a_2]\\\cdots \end{cases}\,. $$ Note that $g$ is well-defined as $(a_n)_n$ is assumed to be monotonically increasing and unconditionally convergent and, evidently, $|f_n|\leq g$ for all $n\in\mathbb N$. Finally, $$ \infty=\int_0^\infty g(x)\,dx=\operatorname{vol}([0,a_1])\cdot\frac1{a_1}+\operatorname{vol}((a_1,a_2])\cdot\frac1{a_2}+\ldots=1+\sum_{n=2}^\infty \frac{a_n-a_{n-1}}{a_n} $$ which concludes the proof.$\quad\square$

Of course the above result is trivial if $\lim_{n\to\infty}\frac{a_{n-1}}{a_n}\neq 1$, but assuming $(a_n-a_{n-1})/a_n\to 0$ as $n\to\infty$, I did not see an easy way to obtain this in an elementary way - admittedly, I didn't think about an alternate proof for too long. Given a quick glance I also did not find a similar result on this site just yet. Hence my question is:

Is there a simpler way to show this result (assuming it holds and I did not make a mistake)? If so, was this maybe already discussed somewhere on math.SE?

Thanks in advance for any answer or comment!

As a final remark (or rather small example), a direct application of this to the sequence $a_n:=\sum_{k=1}^n \frac1k$ immediatly yields the divergence of the series $$ \sum_{n=2}^\infty \frac1{n(\sum_{k=1}^n \frac1k)}=\infty\,. $$ Given the divergence of $\sum_{n=2}^\infty \frac1{n\log(n)}$ this is of course not surprising but nontheless this (at least in my opinion) is interesting to play around with.

Frederik vom Ende
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    Simpler (?) proofs here https://math.stackexchange.com/q/746257/42969 and here https://math.stackexchange.com/q/388898/42969. – Martin R Jan 07 '19 at 13:31
  • @Henry Changed the title to get rid of the ambiguous wording, thank you! – Frederik vom Ende Jan 07 '19 at 13:51
  • @MartinR I'm not sure if the proof in question is directly applicable as the denominator differs from the one used here, i.e. $(a_{n+1}-a_n)/a_n$ vs. $(a_{n+1}-a_n)/a_{n+1}$ (with the latter being smaller so the above result should even be slightly stronger than the linked one?) – Frederik vom Ende Jan 07 '19 at 14:17
  • Using lowercase letters for your sequence, and uppercase letters for the sequences in https://math.stackexchange.com/q/388898/42969, we have $a_n = S_n$, $a_n - a_{n-1} = S_n - S_{n-1} = A_n$, and therefore the identical series$ \sum \frac{a_n-a_{n-1}}{a_n} =\sum \frac{A_n}{S_n}$. Or did I make some error? – Martin R Jan 07 '19 at 14:22
  • Yeah you are of course correct; my annotation rather was related to the post I linked in my comment (which seems to not be equivalent to, but implied by the above result). Anyways, thank you for your time and the useful links! – Frederik vom Ende Jan 07 '19 at 14:33

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Heuristics. Consider its continuum analogue: If $y(t) > 0$ and $y(t) \to \infty$ as $t \to \infty$, then

$$ \int_{0}^{R} \frac{y'(t)}{y(t)} \, \mathrm{d}t = \log y(R) - \log y(0) \xrightarrow[R\to\infty]{} \infty.$$

We can adapt this intuition to our case. Write $b_n = (a_n - a_{n-1})/a_n$. Then it suffices to assume that $b_n \to 0$, for otherwise the conclusion is trivial. Now since

$$ \lim_{n\to\infty} \frac{b_n}{\log a_n - \log a_{n-1}} = \lim_{n\to\infty} \frac{b_n}{-\log(1 - b_n)} = 1, $$

the conclusion follows from

  • (Limit Comparison Test) If $A_n, B_n > 0$ and $\lim A_n/B_n $ converges in $(0, \infty)$, then $\sum_n A_n$ converges if and only if $\sum_n B_n$ converges.

  • $\sum_{n=1}^{N} (\log a_n - \log a_{n-1}) = \log a_N - \log a_0 \to \infty$ as $N\to\infty$, provided $\lim a_n = \infty$.

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Sangchul Lee
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  • A truly splendid argument, thank you! – Frederik vom Ende Jan 07 '19 at 14:27
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    @FrederikvomEnde, Glad it helps! I realized that my argument assumes $b_n > 0$ for all $n$, but this causes no harm as we can remove repetitions from $(a_n)$, i.e., we have $$\sum_{n=2}^{\infty} \left(1 - \frac{a_{n-1}}{a_n} \right) = \sum_{k=1}^{\infty} \left(1 - \frac{a_{n_{k-1}}}{a_{n_k}} \right) $$ where $n = n_k$ is the $k$-th smallest index at which $a_{n-1} < a_n$ holds (with $n_0 := 1$). – Sangchul Lee Jan 07 '19 at 14:33