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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!

Mars Plastic
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dls
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    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 08:18
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    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
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    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. http://dx.doi.org/10.5802/ambp.179 – Martin Sleziak Nov 23 '11 at 16:13
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    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 02:02
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    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
  • This is problem 26 in page 87 of [Stromberg](https://www.amazon.com/Introduction-Classical-Analysis-Chelsea-Publishing/dp/1470425440/ref=sr_1_1?dchild=1&keywords=stromberg+classical+real+analysis&qid=1592946473&sr=8-1). – user2820579 Jun 23 '20 at 21:08
  • I'm wondering if the opposite could also hold. Given positive, decreasing $a_n$, does $na_n\to 0$ imply $\sum a_n$ converges? – Aghbab Jan 05 '22 at 21:43
  • Nope, it doesn't hold! Take $a_n = \frac{1}{n\log n}$. – Aghbab Jan 05 '22 at 21:57

16 Answers16

76

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 $n a_n \to 0.$

Ragib Zaman
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44

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

Aryabhata
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    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 04:13
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    @dls: Right. It is still incomplete though. You have to consider (2n+1)a_{2n+1} too. – Aryabhata Sep 15 '10 at 05:02
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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.

Martin Sleziak
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Pete L. Clark
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    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](http://en.wikipedia.org/wiki/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 07:23
  • @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](http://www.jstor.org/stable/2690790). To add also a freely available paper, it is briefly mentioned [here](http://front.math.ucdavis.edu/1201.5156). (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
  • Can you please update "§2.4.2 of these notes"? It doesn't show anymore. – mike May 09 '22 at 06:51
19

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.

Mars Plastic
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Basilproof
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14

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}+... +a_n \right | < \epsilon$,
$ \Rightarrow \left | a_n + a_{n} + a_{n}+... +a_n \right | \leq \left | a_N + a_{N-1} + a_{N-2}+... +a_n \right | < \epsilon$ $ \Rightarrow \left (n-N)| a_n\right | \leq \left | a_N + a_{N-1} + a_{N-2}+... +a_n \right | < \epsilon$
$ \Rightarrow \left (n-N)| 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 {$a_n$}$\rightarrow 0$, {$na_n$}$\rightarrow 0$ $\blacksquare$

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

kahen
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  • Note that $a_n$ is supposed to be *nonincreasing*. Your sequence isn't. – mrf Sep 12 '13 at 20:01
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    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
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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.

T..
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3

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 $\dfrac \varepsilon 4$.

Michael Hardy
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2

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

MathMan
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    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 04 '16 at 04:27
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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.

Diego
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1

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

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

checkmath
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1

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

1

https://en.wikipedia.org/wiki/Limit_comparison_test

By contradiction if $\lim\limits_{n\to\infty} n a_n\ne 0 $ then $\lim\limits_{n\to\infty} \frac{a_n}{\frac{1}{n}} \ne 0 $ so $\sum\limits_{n=1}^\infty a_n $ and $\sum\limits_{n=1}^\infty \frac{1}{n} $ both have the same status. Knowing that Harmonic series divereges, $\sum\limits_{n=1}^\infty a_n$ diverges as well.

Henry
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1

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 $(a_n)$?

AD - Stop Putin -
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    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
  • @RagibZaman **limit comparison test** has three cases. $$lim_{n\to\infty}\frac{a_n}{\frac{1}{n}}=p$$ Hence $p$ **has to be** either $0$ or $\infty $ or a positive and finite number. Now, you tell me why $p$ can not be $\infty $ or a positive and finite number. – abhishek Feb 09 '22 at 06:08
  • @abhishek $p$ may not exist. – Ragib Zaman Feb 09 '22 at 08:24
1

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 $n a_n \to 0$.

José Siqueira
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0

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.

Aryaman Jal
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  • Sir, Is it true for any convergent series with positive terms? – RIYASUDHEEN TK Dec 07 '19 at 05:26
  • No, this only works when you have add stronger condition that the sequence is monotonically decreasing too. (Note that any sequence of positive terms converging to 0 contains a monotonic subsequence with the same behaviour but a single subsequence is insufficient here.) – Aryaman Jal Dec 07 '19 at 14:48