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

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

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

or

$$\forall \varepsilon > 0 \; \exists n_0 \text{ such that } a_n < \frac{\varepsilon}{n} \; \forall n > n_0$$

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

• This is a classic problem. I remember it from teaching undergrad real analysis in 2005 (specifically I remember having a good think about it at the beautiful YMCA in downtown Montreal -- it took me a while to get it). I think it will be helpful to see a variety of answers. – Pete L. Clark Sep 14 '10 at 8:18
• This is indeed a nice problem, and I don't want to give it away. The hint I would give is to remember to use the condition that $a_n$ is nonincreasing. – David E Speyer Sep 14 '10 at 17:21
• Maybe this is interesting: If we omit monotonicity then $na_n$ converges to 0 statistically. (It is relatively easy to find examples showing that it need not converge in the usual sense.) Tibor Šalát; Vladimír Toma: A Classical Olivier’s Theorem and Statistical Convergence. dx.doi.org/10.5802/ambp.179 – Martin Sleziak Nov 23 '11 at 16:13
• Maybe it's too late for your exam, but I've added another answer that is fairly short and is not very notation-intensive and does not rely on standard results but works from basic definitions. (I don't even use the fact that the harmonic series diverges.) I like to keep things simple. $\qquad$ – Michael Hardy Jul 21 '16 at 2:02
• Can you tell me where did you see this problem? I'm studying for an exam and It would really help some extra problems. This seems a very good one – J.Dane Jul 27 '18 at 10:39

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 $S_{2n} - S_{n}$?

• Let me give this a shot: Assuming the limit you mention tends to zero, for a given $\epsilon$, there is an integer $n_0$ such that $2 ( S_{2n_0} - S_{n_0} ) < \epsilon$. Because the sequence is nonincreasing, we have that $2n a_{2n} \leq 2 ( S_{2n_0} - S_{n_0} ) < \epsilon$ for all $n \geq n_0$. – dls Sep 15 '10 at 4:13
• @dls: Right. It is still incomplete though. You have to consider (2n+1)a_{2n+1} too. – Aryabhata Sep 15 '10 at 5:02

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 $n a_n \to 0.$

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.

• Re: P.S.: The English Wikipedia page is almost a verbatim copy of the German one. The Pringsheim family is very well known in Germany due to its influence in Munich over centuries, and also due to fact that Alfred P.'s daughter was married to Thomas Mann who certainly contributed to making the family's fate in the first half of the twentieth century known to a wide audience. – t.b. Nov 23 '11 at 7:23
• @t.b.: Thanks for this. I vaguely noticed the connection to Thomas Mann in the wikipedia article, but it didn't really sink in. – Pete L. Clark Nov 23 '11 at 16:04
• Interestingly, it seems that Olivier's paper contained a mistake and later Abel wrote a paper to correct it. Details can be found e.g. in Michael Goar: Olivier and Abel on Series Convergence, Mathematics Magazine Vol. 72, No. 5 (Dec., 1999), pp. 347-355 jstor. To add also a freely available paper, it is briefly mentioned here. (The later is the paper where I've read about Olivier's mistake for the first time.) – Martin Sleziak Aug 15 '12 at 11:07

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

Because the series converges the sequence of partial sums forms a Cauchy sequence so we have for m,n>N

$$\| \sum_m^n(a_i) \| < \epsilon$$

And by a similar argument to those above: $\|(n-m)a_n \| \leq \| \sum_m^n(a_i) \| < \epsilon$ Distributing out the left hand side and spliting the inequality gives us the chain:

$$\| na_n \|-\|ma_n\|\leq \|(n-m)a_n \| \leq \| \sum_m^n(a_i) \| < \epsilon$$

Then in the limit as n goes to infinity and fixing m we have the desired result since $ma_n$ will go to zero.

I think... ?

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.

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
$\displaystyle\qquad \sum_{n=1}^\infty a_n = \sum_{n=1}^\infty \frac{S(n)}n = \sum_{k=1}^\infty \frac1{k^2} < \infty$.

• Note that $a_n$ is supposed to be nonincreasing. Your sequence isn't. – mrf Sep 12 '13 at 20:01
• Ah. Miss one word and the whole thing goes down the crapper. Typical. Well now it's fixed at least. And CW just for good measure. – kahen Sep 12 '13 at 20:02

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 {$a_n$}$\rightarrow 0$, {$na_n$}$\rightarrow 0$ $\blacksquare$

• Shouldn't it be $|a_N + a_{N+1} + \ldots + a_n|$ since $n \ge N$? – philmcole Jan 13 '18 at 10:07
• Very clear and nicely explained. Thanks – StammeringMathematician Jul 28 at 18:31

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. $\lim a_n=0 \implies \lim_{n \rightarrow \infty} n a_n = 0$

• You seem to claim that directly from $(n-m)a_n<\varepsilon$ you get $(n-m)a_n=0$. This argument is not correct, since $m$ depends on $\varepsilon$. (But you still get $(n-m)a_n<\varepsilon$. For a fixed $\varepsilon>0$ and for $m$ corresponding to this $\varepsilon$.) – Martin Sleziak Apr 4 '16 at 4:27

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 $(a_n)$?

• I can't see how this works. The easy part - if the limit exists and is positive, then $\sum a_n$ diverges like a harmonic series, contradiction. The harder part, if the limit does not exist - There there exists a subsequence and an $\epsilon > 0$ such that $n_k a_{n_k} > \epsilon$ for all $n_k$. But that itself does not imply the result. Did you have something else in mind? – Ragib Zaman Oct 11 '11 at 15:03

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 $n a_n \to 0$.

• Consider $a_n = \frac{1}{n^2}$. Then the limit of ratios is $1$. – Daniel Fischer Apr 27 '14 at 21:59
• Oh my, you are right. Nevermind then. – José Siqueira Apr 27 '14 at 22:06
• Still, looking at a series is a nice trick when you want to prove a limit is 0. – José Siqueira Apr 27 '14 at 22:10

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 $\dfrac \varepsilon 4$.

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.

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

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

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, $$\displaystyle \lim_{n\to \infty}nf(a_n) =0$$ and we are done.