Series converges implies $\lim{n a_n} = 0$

Question

I'm studying for qualifying exams and ran into this problem.

Show that if ${a_n}$ is a nonincreasing sequence of positive real numbers such that $\sum_n{a_n}$ converges, then $\lim_{n \rightarrow \infty}{n a_n} = 0$.

Using the definition of the limit, this is equivalent to showing

\begin{equation} \forall \epsilon > 0 \; \exists n_0 \; \mbox{such that} \; |n a_n| < \epsilon \; \forall n > n_0 \end{equation}

or

\begin{equation} \forall \epsilon > 0 \; \exists n_0 \; \mbox{such that} \; a_n < \frac{\epsilon}{n} \; \forall n > n_0 \end{equation}

Basically, the terms must be bounded by the harmonic series. Thanks, I'm really stuck on this seemingly simple problem!

Answer 1

Some hints:

If $S_{n} = \sum_{k=1}^{n} a_{k}$

then what is

$\lim_{n \to \infty} S_{2n} - S_{n}$?

Now can you use the fact that $a_{n}$ is non-increasing to upper bound a certain term of the sequence $na_{n}$ with a multiple of $S_{2n} - S_{n}$?

Answer 2

By the Cauchy Condensation test, $\displaystyle \sum 2^n a_{2^n} $ converges since $\displaystyle \sum a_n $ converges, and thus $ 2^n a_{2^n} \to 0. $ Now for $ 2^n < k < 2^{n+1} $,

$$ 2^n a_{2^{n+1}} \leq k a_{k} \leq 2^{n+1} a_{2^n}$$

so $n a_n \to 0.$

Answer 3

Now that enough time has passed so that more information will not spoil anything for the OP:

This fact can be found in $\S 179$ of G.H. Hardy's seminal A Course of Pure Mathematics: he mentions that it was first proved by Abel, then forgotten and later rediscovered by Alfred Pringsheim. I have reproduced Hardy's proof in $\S 2.4.2$ of these notes on infinite series. This is much slicker than what I came up with when I had to solve this exercise myself some years ago. On the other hand it seems to be exactly what Aryabhata's answer hints at.

In my notes I also attribute this result to L. Olivier and even cite the issue of Crelle's Journal in which it appears in 1827. This attribution does not appear in Hardy's book, which temporarily mystified me (I am no historian of mathematics: whatever such information I have comes from math books with good bibliographies), but I surmise I must have gotten it from Konrad Knopp's book on infinite series (the only other book I own which treats the subject seriously).

P.S.: The wikipedia article on Pringsheim is unusually (almost suspiciously?) good. The impression that I have of him as a mathematician is someone who worked on infinite series at a stage when the foundations of the theory were finally solidly in place...and when the best mathematicians of the day had gone on to more fundamental and difficult problems. But I don't know whether this is at all fair. Anyway, it seems that you won't hear of him until you learn a little more about series than is treated in the standard contemporary curriculum, but as soon as you do his name comes up again and again.

Answer 4

Try to stay away from quantifier-laden formulas, which make the problem harder to understand, and draw a picture. There is an equivalent problem for decreasing functions (as in the Integral Test for convergence) and the picture makes it obvious what is true in that case. Having seen the continuous proof, run the same argument for the sequence, or specialize the function to a sequence by using step functions or approximation thereof. I won't spoil the "aha!" proof-by-picture experience by posting more details, but it is quite easy once you draw the graph.

Answer 5

You might also do this the other way around. What if $$\lim_{n\to\infty}na_n \not=0,$$ that is if the limit does not exist or if it is positive, what does this tell you about $a_n$? What about sub-sequences of $(a_n)$?