9
$\begingroup$

I would like to know the relationships between bounded and convergent series. By bounded series I mean a series whose sequence of partial sums is bounded. For example, it seems natural that if a series is convergent, it is also bounded, but does the converse hold?

Thanks in advance,

$\endgroup$
7
  • $\begingroup$ "if a series is convergent, it is also convergent". True... but I don't think that's what you meant. One of those "convergent" was meant to be "bounded." Which one? $\endgroup$ – Arturo Magidin Sep 6 '11 at 20:03
  • 3
    $\begingroup$ You presumably mean if it is convergent, its partial sums are bounded. That's true. The converse does not hold. Consider for example $1-1+1-1+\cdots$. But if the terms are positive, and the partial sums are bounded, then the series converges. $\endgroup$ – André Nicolas Sep 6 '11 at 20:04
  • $\begingroup$ Davide: by bounded I mean that the sequence of partial sums is bounded, as in en.wikipedia.org/wiki/Bounded_function $\endgroup$ – Federico Sep 6 '11 at 20:08
  • $\begingroup$ @Federico Ah. It may be better spell that out, rather than saying "bounded series". $\endgroup$ – Dylan Moreland Sep 6 '11 at 20:19
  • 1
    $\begingroup$ But there is one important special case when there's a tight relation between bounded partial sums and convergence. Suppose that the series $\sum_n x_n$ consists of only nonnegative terms. (E.g., $\sum 1/n$ or $\sum 1/n^2$.) Then the series is convergent if and only if its sequence of partial sums $\sum_{i=1}^n x_i$ is bounded. $\endgroup$ – Srivatsan Sep 6 '11 at 20:45
27
$\begingroup$

Whenever we have a series, $$\sum_{i=1}^{\infty} a_i,$$ we "automatically" get two sequences out of that series:

  1. The sequence of terms, which is $a_1,a_2,a_3,\ldots$; and
  2. The sequence of partial sums, which is $s_1,s_2,s_3,\ldots$, where $$\begin{align*} s_1 &= a_1\\ s_2 &= a_1+a_2\\ s_3 &= a_1+a_2+a_3\\ &\vdots\\ s_n &= \sum_{i=1}^n a_i = a_1+a_2+\cdots + a_n. \end{align*}$$

When we talk about "convergence of the series", we are really talking about convergence of the sequence of partial sums: the series $\sum a_i$ converges if and only if the sequence $(s_n)$ converges. That is, your definitions about "series" are really about "sequence of partial sums", and so you have the usual relationship:

In particular, $$\sum_{i=1}^{\infty}a_i\text{ converges}\Longleftrightarrow \{s_i\}_{i=1}^{\infty}\text{ converges}\Longrightarrow \{s_i\}_{i=1}^{\infty}\text{ is bounded}\Longleftrightarrow \sum_{i=1}^{\infty}a_i\text{ is bounded}$$ (where "is bounded" is as per your definition above); but it is possible for $\{s_i\}_{i=1}^{\infty}$ to be bounded, and not convergent, so one can have a series $\sum_{i=1}^{\infty}a_i$ that is bounded (i.e., the sequence of partial sums is bounded) but does not converge.

A simple example of this is $\sum_{i=1}^{\infty} (-1)^n$. The partial sums are $s_{2k+1} = -1$ and $s_{2k}=0$ for every $k$, so the sequence of partial sums is: $$-1,\ 0,\ -1,\ 0,\ -1,\ldots$$ which is bounded but not convergent. So the series is bounded but not convergent.

The relevant theorem for sequences, as you are no doubt aware, is:

Theorem. If $\{b_n\}$ is a monotone sequence, then $\{b_n\}$ converges if and only if it is bounded.

How does that translate for series? When is the sequence of partial sums monotone?

$\{s_i\}$ is increasing if and only if $s_n\leq s_{n+1}$ for all $n$, if and only if $s_{n+1}-s_n\geq 0$ for all $n$; but $s_{n+1}-s_n = a_{n+1}$. So:

The sequence of partial sums of $\displaystyle \sum_{i=1}^{\infty}a_i$ is increasing if and only if all the terms $a_i$ are nonnegative. The sequence of partials sums is strictly increasing if and only if all the terms $a_i$ are positive.

Likewise,

The sequence of partial sums of $\displaystyle \sum_{i=1}^{\infty}a_i$ is decreasing if and only if all the terms $a_i$ are nonpositive. The sequence of partial sums is strictly decreasing if and only if all the terms $a_i$ are negative.

So we conclude:

Theorem. Let $\displaystyle \sum_{i=1}^{\infty}a_i$ is a series in which every term $a_i$ is nonnegative. Then the series converges if and only if it is bounded (in the sense that the sequence of partial sums is bounded).

$\endgroup$
1
  • $\begingroup$ You said it is possible that one can have a series that is bounded (i.e., the sequence of partial sums is bounded) but does not converge. Is this also true for the absolutely convergent case? So is it possible to have a sequence of partial sums that is bounded, but where the associated series does not converge absolutely? $\endgroup$ – Kamil Jul 21 '16 at 10:40
4
$\begingroup$

No, a bounded series does not necessarily converge. Consider the series $\displaystyle \sum (-1)^n $ (heavily related to Henning's example). It will forever oscillate between 0 and 1 (or -1 and 0, depending on the indices).

But if the partial sums are bounded and monotonic, then it does converge.

But in either case, it's a bit weaker than the converse - convergent series always have bounded partial sums.

$\endgroup$
7
  • $\begingroup$ It may happen that the partial sums of a series are bounded and its general term converges to zero and yet the series diverges. Example: if $a_n=(-1)^k2^{-k}$ for every nonnegative $k$ and $n$ such that $2^k\le n<2^{k+1}$, then $a_n\to0$ but the limit set of the sequence of the partial sums $\sum a_n$ is the interval $[a_0,a_0+1]$. $\endgroup$ – Did Sep 6 '11 at 20:32
  • $\begingroup$ @Didier: That's a great counterexample. I'll hold onto that. $\endgroup$ – davidlowryduda Sep 6 '11 at 22:25
  • $\begingroup$ @Didier: One more thing: are there a weaker set of conditions than bounded and monotonic? (I know this is very open, but perhaps there is some 'surprising' or 'deceptive' set of conditions?). $\endgroup$ – davidlowryduda Sep 6 '11 at 22:26
  • $\begingroup$ Well yes: absolutely convergent. $\endgroup$ – Did Sep 6 '11 at 23:56
  • $\begingroup$ @Didier: I don't understand: are you saying it's surprising that something that converges absolutely also converges in general? $\endgroup$ – davidlowryduda Sep 7 '11 at 0:11
2
$\begingroup$

A convergent sequence is bounded, but a bounded sequence is not necessarily convergent. Consider, for example the sequence (1, -1, 1, -1, 1, -1, ...).

On the other hand, an increasing (or decreasing) bounded sequence in $\mathbb R$ will necessarily converge.

$\endgroup$
5
  • $\begingroup$ I know about that about sequences, but what about series? Does it even make sense to say "bounded series"? $\endgroup$ – Federico Sep 6 '11 at 20:04
  • $\begingroup$ I assumed you meant sequences because "bounded series" is not immediately meaningful to me -- except if it means that the sequence of partial sums is bounded (in which case boundedness and convergence of the series are both the same as the sequence, so the relation is the same). $\endgroup$ – hmakholm left over Monica Sep 6 '11 at 20:05
  • 3
    $\begingroup$ @Federico: When talking about a series, we usually refer to the "sequence of partial sums"; in particular, a series converges if and only if the sequence of partial sums converges, and a series is bounded if and only if the sequence of partial sums is bounded. $\endgroup$ – Arturo Magidin Sep 6 '11 at 20:14
  • $\begingroup$ @Henning: yes, I was referring to the sequence of the partial sums. I find a conflict between what you say and what André Nicolas says, because you impose the condition of being an increasing (decreasing) bounded sequence and he imposes the condition of being a bounded sequence with all the terms positive to converge. Could you explain? $\endgroup$ – Federico Sep 6 '11 at 20:22
  • 1
    $\begingroup$ @Federico: if all terms of the series are positive, then the sequence of partial sums is increasing; if all terms of the series are negative, then the sequence of partial sums is decreasing. $\endgroup$ – Arturo Magidin Sep 6 '11 at 20:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.