If $(a_n)\subset[0,\infty)$ is non-increasing and $\sum a_n<\infty$, then $\lim{n a_n} = 0$

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 \varepsilon > 0 \; \exists n_0 \text{ such that } |n a_n| < \varepsilon \; \forall n > n_0 \end{equation}

or

\begin{equation} \forall \varepsilon > 0 \; \exists n_0 \text{ such that } a_n < \frac{\varepsilon}{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 so $ 2^n a_{2^n} \to 0. $ 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

I think I have an answer which doesn't rely on being clever enough to use the even and odd subsequences.

Since the series converges, the sequence of partial sums forms a Cauchy sequence. Hence, for all $\epsilon>0$ there is an $n\in\Bbb N$ sucht that for all $n>m>N$ we have

$$ \sum_{m+1}^n a_k < \epsilon. $$

Due to the monotonitcy of $a_n$, we also have

$$ (n-m)a_n\le\sum_{m+1}^n a_k, $$

and combining the two previous inequalities leads to

$$ na_n \leq \epsilon+ma_n. $$

Since $a_n$ goes to zero, this yields $$\limsup_{n\to\infty}na_n\le\epsilon,$$ and as $\epsilon$ was arbitrary, the claim follows.

Answer 5

Let $\epsilon > 0$.

Let Let $s_n=\sum_{k=1}^n a_k$.

$\{s_n\}$ converges, therefore $\{s_n\}$ is Cauchy which implies $\exists$ some $N$ such that $m, n \geq N $ implies $\left | s_n-s_m \right | < \epsilon$

Consider $m=N$, for $n \geq N$ we have $\left | s_n-s_N \right | < \epsilon$,
$ \Rightarrow \left | a_N + a_{N-1} + a_{N-2}+\cdots +a_n \right | < \epsilon$,
$ \Rightarrow \left | a_n + a_{n} + a_{n}+\cdots +a_n \right | \leq \left | a_N + a_{N-1} + a_{N-2}+\cdots +a_n \right | < \epsilon$ $ \Rightarrow \left (n-N)| a_n\right | \leq \left | a_N + a_{N-1} + a_{N-2}+\cdots +a_n \right | < \epsilon$
$ \Rightarrow (n-N)\left| a_n\right | < \epsilon$

So, $\lim_{n \rightarrow \infty } (n-N)a_n = 0$
$\lim_{n \rightarrow \infty } na_n - \lim_{n \rightarrow \infty } Na_n = 0$
From above, since {$a_n$}$\rightarrow 0$, {$na_n$}$\rightarrow 0$ $\blacksquare$

Answer 6

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 7

Note that the desired conclusion is not true if the $a_n$ are not assumed to be nonincreasing:

We know a couple of facts:

1. If a sequence $(s_n)_{n\geq1}$ converges then we have that
   $\displaystyle\qquad\liminf_{n\to\infty} s_n = \lim_{n\to\infty} s_n = \limsup_{n\to\infty} s_n$.

2. There is a summable series $\sum_{n=1}^\infty a_n$ with non-negative $a_n$ such that $\limsup\limits_{n\to\infty} na_n > 0$.

Take these two together you get that the best you can hope to prove in this slightly more general situation is that $\liminf\limits_{n\to\infty} na_n = 0$, and that is indeed true.

For an example showing that 2 above is true, consider this: We start with the harmonic series and observe that
$\displaystyle\qquad \limsup_{n\to\infty} n\cdot \tfrac1n = 1$,
but it isn't summable. But we could replace infinitely many of the sequence elements by zeroes without changing the limit superior from being $1$. Define $S(n)$ to be $1$ when $n$ is a perfect square and $0$ otherwise. Then $a_n = \frac{S(n)}n$ has the desired property as
$\displaystyle\qquad \sum_{n=1}^\infty a_n = \sum_{n=1}^\infty \frac{S(n)}n = \sum_{k=1}^\infty \frac1{k^2} < \infty$.

Answer 8

If $\lim\limits_{n\to\infty} n a_n\ne 0$, then for some $\varepsilon>0$, there are infinitely many $n$ for which $a_n \ge \dfrac \varepsilon n.$

If $a_n \ge \dfrac \varepsilon n,$ then let $m = \lfloor \log_2 n \rfloor$, so that $2^m\le n < 2^{m+1}$. Then $$ a_{2^{m-1}} + \cdots + a_{2^m} \ge 2^{m-1} a_{2^m} \ge 2^{m-1} a_n \ge \frac{2^{m-1}\varepsilon} n \ge \frac{2^{m-1}\varepsilon} {2^{m+1}} = \frac \varepsilon 4. $$

Then go on to the next $n$ for which $a_n \ge \dfrac \varepsilon n$, and if it's not big enough for the resulting sequence $a_{2^{m-1}} + \cdots + a_{2^m}$ to have a different (larger) value of $m$ than the one we just saw, then go on to the next one after that, etc.

We get infinitely many disjoint sets of terms with sums exceeding $\dfrac \varepsilon 4$.

Answer 9

Here's one more approach : By Cauchy's general principle of convergence :

$\forall \epsilon>0, \exists ~p \in \mathbb N , \exists ~m> N: |a_{m+1}+a_{m+2}+ \cdots + a_{m+p}| < \epsilon$.

Choose:$ p = n-m$ .

$|a_{m+1}+a_{m+2}+ \cdots + a_n| < \epsilon$.

Now, since $a_n$ is non increasing and $\epsilon \rightarrow 0$: the above relation can be written as :

$(n-m)a_n = 0 \implies na_n = ma_n $

For a convergent series. $\lim a_n=0 \implies \lim_{n \rightarrow \infty} n a_n = 0$

Answer 10

If the sequence is strictly decreasing, I guess there is a neat little argument. Look at the series $\sum n a_n$ and define $\rho_n=\frac{(n+1)a_{n+1}}{na_n}$. Then $\rho_n=\frac{a_{n+1}}{a_n} \frac{(n+1)}{n}$. We have that $\lim_{n \to \infty} \frac{(n+1)}{n}$ exists and is equal to $1$. As $(a_n)$ is strictly decreasing and the terms are positive, we have $\frac{a_{n+1}}{a_n} < 1$ Therefore the limit $\lim_{n \to \infty} \frac{a_{n+1}}{a_n}$ exists and is smaller than $1$. So $\lim_{n \to \infty} \rho_n$ exists and is smaller than 1, and thus by the ratio test the series $\sum n a_n$ converges. Hence $n a_n \to 0$.

Answer 11

Let $\epsilon$ be given. By Cauchy, there exists $N(\epsilon)$ in $\mathbb{N}$ such that: $a_N+a_{N+1}+....+a_{N+P}<\epsilon$, for every $P$ in $\mathbb{N}$. Note that for every chosen $P$, there are $P$ $a_n$'s on the left side of the inequality, not all of which can exceed $\dfrac{\epsilon}{P}$. Anyway, since $a_n$ is non increasing, $a_{N+P}$ is always less than $\dfrac{\epsilon}{P}$. We have then:

$$a_{N+P}<\dfrac{\epsilon}{P}, \forall P \in \mathbb{N}$$ Multiplying by $N+P$ at each side of the inequality: $$(N+P)a_{N+P}<\dfrac{(N+P)\epsilon}{P}$$

"Throwing" $p$ to ininity, we can see then that for $n$ large enough, $na_n$ is almost less than $\epsilon$.

Taking $\dfrac{\epsilon}{2}$ instead of $\epsilon$, we can guarantee that $na_n$ is less than $\epsilon$ for $n$ large enough, completing the proof.

Answer 12

Suppose $n_ja_{n_j}\geq \alpha>0$ and up to (infinity) subsequence assume $n_{j}-n_{j-1}\geq n_{j}/2$.

Using that $a_n\geq 0$ we get

$$\displaystyle A=\sum_{n=1}^\infty a_n =\sum_{n=1}^{n_0} a_n+\sum_{j=1}^\infty\sum_{n=n_{j-1}+1}^{n_j}a_n\geq B+\sum_{j=1}^\infty\sum_{n=n_{j-1}+1}^{n_j}a_{n_j}\geq B+\sum_{j=1}^\infty(n_{j}-n_{j-1})a_{n_j} \\\geq B+\sum_{j=1}^\infty\frac{n_{j}}{2}a_{n_j}\geq B+\frac{1}{2}\sum_{j=1}^\infty{n_{j}}{}a_{n_j}=\infty. \ QED. $$

Answer 13

There is also quite nice proof using Stolz-Cesaro Theorem: $\lim_{n\rightarrow\infty}a_n=\lim_{n\rightarrow\infty}a_{n+1}=0$ because of series convergence.

$$ \lim_{n\rightarrow\infty}\sum_{k=1}^n a_k=g $$ Using Stolz-Cesaro Theorem: $$ g=\lim_{n\rightarrow\infty}\frac{n \sum_{k=1}^n a_k}{n} = \left[\frac{\infty}{\infty} \right] = \lim_{n\rightarrow\infty}\frac{(n+1) \sum_{k=1}^{n+1} a_k-n\sum_{k=1}^n a_k}{n+1-n} = \lim_{n\rightarrow\infty}(\sum_{k=1}^n a_k + (n+1)a_{n+1}) $$ $$ g=g+\lim_{n\rightarrow\infty}a_{n+1}+\lim_{n\rightarrow\infty}na_{n+1} $$ $$ \lim_{n\rightarrow\infty}na_{n+1} = \lim_{n\rightarrow\infty}na_{n}=0 $$ $\square$

Answer 14

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 $(a_n)$?

Answer 15

Let $f: (0, \infty)\to [0, \infty)$ be a continuous, non-increasing function such that $f(n)=a_n$ for all $n \in \mathbb{N}.$ Such a function certainly exists (for $x \in (n, n+1)$ let $f(x) = a_n + (a_{n+1}-a_n)(x-n)$)

Since $\displaystyle \sum_{n=1}^{\infty}f(n) = \displaystyle \sum_{n=1}^{\infty} a_n$ converges, by the integral test, $\displaystyle \int_{0}^{\infty}f(x)dx < \infty.$

Claim: $\displaystyle\lim_{x\to \infty}xf(x)=0.$

We reproduce the argument that JLA gave here, begin by noting that $\displaystyle \lim_{t \to \infty}\int_{t/2}^{t}f(x)dx=0.$ Then $0 \leq \dfrac{t f(t)}{2}\leq \displaystyle \int_{t/2}^{t}f(x)dx \to 0$ as $t \to \infty$ and the claim holds.

In particular, $\displaystyle \lim_{n\to \infty}na_{n} =0$ and we are done.