A convergent sequence $\{a_n\}$ and divergent sequence $\{b_n\}$ such that $\{a_n+b_n\}$ is convergent

Question

Give an example of a convergent sequence $\{a_n\}$ and divergent sequence $\{b_n\}$ such that $\{a_n+b_n\}$ is a convergent series.

I've been trying to solve this question for a couple days now and have been struggling, if anyone could give me a hint or show me how you got your answer as I feel this isn't solvable but the question says that I must have an example. Thank you in advance, Math Student :)

Who asked you that? You're unlikely to find an example, it is not possible. — Clement C., Jan 23 '17 at 21:32
As pointed out by Nayuki, your title and body are asking subtly but crucially different questions. Is the sum supposed to be a convergent sequence, or a convergent series? (And for that matter, what about ${a_n}$ and ${b_n}$?) — user2357112, Jan 24 '17 at 03:31
You still need to decide whether you mean "sequence" or "series"; I merely improved your LaTeX and title. — user21820, Jan 24 '17 at 07:40

score 14 · Answer 1 · answered Jan 23 '17 at 21:28

14

Assume ${a_n + b_n}$ converges. Since ${a_n}$ converges, ${a_n + b_n - a_n}$ converges, contradicting the fact that ${b_n}$ does not converge.

answered Jan 23 '17 at 21:28

5

Does your answer take into account the fact that in the question, $a_n$ is a convergent sequence, $b_n$ is a divergent sequence, and $a_n+b_n$ is a convergent series? – Nayuki Jan 24 '17 at 03:06
2

@Nayuki If by convergent series you mean that the series over $a_n+b_n$, i.e. $\sum_{i=1}^\infty (a_i+b_i)$, converges, then this implies that the sequence $(a_n+b_n)$ converges. – Hermann Döppes Jan 24 '17 at 05:55
{$a_n$},{$b_n$}, and {$a_n$+$b_n$} are all sequences, sorry for leaving that out – user2956671 Jan 24 '17 at 22:52

score 4 · Answer 2 · answered Jan 23 '17 at 21:43

This is not possible. Suppose it was, then we have a convergent sequence ${a_n+b_n}$, and ${a_n}$. We know that the difference of convergent sequences is itself, convergent. This means if ${a_n+b_n}$ converges, and ${a_n}$ converges, then ${a_n+b_n-a_n}$ converges, but this means that ${b_n}$ is convergent, which contradicts our hypothesis, so no, this cannot be done.

You can however have two divergent sequences sum to a convergent one. Just take ${a_n}=(n)$ and ${b_n}=(-n)$ which gives us ${a_n+b_n}=(0)$, and the zero sequence is a constant sequence, which is trivially convergent.

score 0 · Answer 3 · answered Mar 25 '20 at 18:57

Since you were seeking an example and one could not be found, but neither answer has been accepted, I can offer this as an example of one... although it may not be appropriate in the context of the tag real analysis.

Let $\forall i:a_i,b_i\in\Bbb Z[\frac12]/\Bbb Z$ the dyadic rationals in the interval $(0,1]$

Then consider the sequences:

The convergent sequence $b_n=\frac12,\frac34,\frac78,\frac{15}{16}\ldots\to1$

and the divergent sequence $1-b_n$

Their sum is the constant sequence $1,1,1\ldots$ which converges in $(0,1]$.

A convergent sequence $\{a_n\}$ and divergent sequence $\{b_n\}$ such that $\{a_n+b_n\}$ is convergent

3 Answers3