True or false: If $\{a_n\}$ is divergent, then the series summation of $\sum_{n=1}^\infty (a_{n+1}-a_n$) is divergent.

I know that this can be split into two summations: $$\sum_{n=1}^\infty a_{n+1}-\sum_{n=1}^\infty a_n$$

Earlier I learned that if $\{a_n\}$ and $\{b_n\}$ are divergent, then it is false to say that $\{a_n+b_n\}$ is also divergent because $a_n$ could equal $n$ and $b_n$ could equal $-n$, in which case they add to zero, which converges. How do these two situations differ?

How could one be sure that they aren't subtracting away whatever is making the sequence divergent in the first place? What does it mean to subtract the $n$th term from the $(n+1)$-th term in a sequence?

  • $\begingroup$ Oh come on, this is the second time today I've been too slow on edits... $\endgroup$ – teadawg1337 Dec 14 '14 at 23:25
  • $\begingroup$ @teadawg1337 : You attempted to change $\{a_n\}$ to $a_n$. I think that is a mistake, since $a_n$ is often taken to mean the $n$th term, whereas the notation with braces is taken to mean the whole sequence. $\endgroup$ – Michael Hardy Dec 14 '14 at 23:34
  • $\begingroup$ @MichaelHardy It was my intention to put \{\} around them, I must've forgotten $\endgroup$ – teadawg1337 Dec 14 '14 at 23:35

The identity $$ \sum_{n=1} (a_{n+1}-a_n)=\sum_{n=1}^\infty a_{n+1} - \sum_{n=1}^\infty a_n $$ is true if both of the sums on the right converge. But it sometimes fails when they don't. For example, suppose $a_n=1+\dfrac 1 {2^n}$. Then both of the series on the right diverge, but the one on the left converges.

The sum $\sum (a_{n+1}-a_n)$ "telescopes", i.e. $$ \sum_{n=1}^N (a_{n+1}-a_n) = (a_2-a_1)+(a_3-a_2)+(a_4-a_3)+\cdots+(a_{N+1}-a_N) $$ and all of the terms cancel except the first and last ones, so the sum is $a_{N+1}-a_1$.

Maybe the cancelation is clearer if it's written like this: \begin{align} & (-a_1+\underbrace{a_2)+(-a_2}+\underbrace{a_3)+(-a_3}+\underbrace{a_4)+{}\quad}\cdots\underbrace{\quad{}+(-a_{N-1}}+\underbrace{a_N)+(-a_N}+a_{N+1}) \end{align}


True, because partial sum $S_n = \displaystyle \sum_{k=1}^n \left(a_{k+1}-a_k\right) = a_{n+1} - a_1 \to \infty$

  • 1
    $\begingroup$ It is not guaranteed to diverge to infinity, however it will not converge. Consider the sequence $a = \{-1,1,\dots,(-1)^n,\dots\}$. You have then that $a_{n+1}-a_1 = -1 \pm 1$ $\endgroup$ – JMoravitz Dec 14 '14 at 23:30
  • $\begingroup$ How do you know that the summation equals a_n+1 - a_1? $\endgroup$ – MaryG Dec 14 '14 at 23:32
  • $\begingroup$ @Anonymous Note the cancellation in $(a_4-a_3)+(a_3-a_2)+(a_2-a_1)=a_4-a_1$ $\endgroup$ – Milo Brandt Dec 14 '14 at 23:33
  • $\begingroup$ Why doesn't the series a_n+1 include the term a_1? $\endgroup$ – MaryG Dec 14 '14 at 23:38

Despite several other answers, I feel compelled to add my own as I disagree with all appearing so far.

There are two interpretations: Either the problem statement is saying that the sequence $\{a\}_n$ doesn't converge. The other interpretation is that the series $\sum_{n=1}^\infty a_n$ doesn't converge.

In the first interpretation, as the sequence $\{a\}_n$ doesn't converge, then it fails the cauchy criterion and there is some $\epsilon>0$ such that there are infinitely many occurences of $|a_{n+1} - a_n|>\epsilon$. As such, the series $\sum_{n=1}^\infty b_n = \sum_{n=1}^\infty (a_{n+1} - a_n)$ will diverge due to the fact that the terms $b_n$ do not converge to zero. As such, the claim is true and the series is divergent.

Theorem: If a series $\sum_{n=1}^\infty b_n$ is convergent $\Rightarrow$ the terms $b_n$ converge to zero as $n\to \infty$. The contrapositive is then if the terms $b_n$ do not converge to zero, then the series is divergent.

In the second interpretation, if the series $\sum_{n=1}^\infty a_n$ diverges, you can in fact have a convergent sum for $\sum_{n=1}^\infty b_n = \sum_{n=1}^\infty (a_{n+1} - a_n)$ in the case that $\{a\}_n = (1,1,1,1,\dots)$ as $a_{n+1} - a_n = 0$ for all $n$. As such, the response to the claim would be "false" since it is "often but not always true that it will diverge"

  • $\begingroup$ What do you disagree with in my answer? $\endgroup$ – Michael Hardy Dec 14 '14 at 23:53
  • $\begingroup$ @MichaelHardy Your answer has a great deal of good information, and at the time of writing that line there was another one or two answers which have since been deleted. However, I do think it makes an assumption as to the notation of the problem writer, assuming that it is the second interpretation I listed above, which I hold may not necessarily be the case. The punchline as to the answer to the true/false statement is also missing, and may be missed by the reader. $\endgroup$ – JMoravitz Dec 14 '14 at 23:58
  • $\begingroup$ I'm leaving that last part as an excercise. There's no need for the Cauchy criterion unless that is needed to prove that $\lim_N (a_N-a_1)$ does not exist if $\lim_N a_N$ does not exist. Your second interpretation seems implausible to me. I think one would write "$\sum_n a_n$ diverges" rather than "$\{a_n\}$ diverges" if the former were what was meant. ${}\qquad{}$ $\endgroup$ – Michael Hardy Dec 15 '14 at 0:04
  • $\begingroup$ That's a lot of fuzz around something quite simple. The statement which the OP is refering to is: Let $(a_n)$ be a sequence, $v_n = a_{n+1} - a_{n}$ and $S_n = \sum_{k=1}^n v_k$. Then $(a_n)$ and $(S_n)$ are of the same nature, i.e: $(a_n)$ converges iif $(S_n)$ converges. Of course the contrapositive gives: $(a_n)$ diverges iif $(S_n)$ diverges. And the reason why this is true is because: $S_n = a_{n+1} - a_1 $ by telescoping. Of course since the series $\sum v_n$ converges iif the sequence $(S_n)$ converges, you get the OP's proposition $\endgroup$ – mvggz Dec 15 '14 at 13:09

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