A convergent sequence $\{a_n\}$ and divergent sequence $\{b_n\}$ such that $\{a_n+b_n\}$ is convergent 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 :)  
 A: 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. 
A: 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.
A: 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 $(0,1]$.
