Divergence of Summation $a_{n+1} - a_n$? 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?
 A: 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}
A: True, because partial sum $S_n = \displaystyle \sum_{k=1}^n \left(a_{k+1}-a_k\right) = a_{n+1} - a_1 \to \infty$
A: 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"
