Question: I have two null sequences $(a\_n)\_{n\in\mathbb N}$ and $(b\_n)\_{n\in\mathbb N}$ with $a\_n \ge 0$ and $b\_n \le 0$ for all $n\in \mathbb N$. Let $(c\_n)\_{n\in\mathbb N}$ be a sequence with either $c\_n=a\_n$ or $c\_n=b\_n$. I do not know which member of $c\_n$ is actually $a\_n$ and which is $b\_n$. I know, that until the n-th member $k(n)$ members of $(c\_n)\_{n\in\mathbb N}$ are from $(a\_n)\_{n\in\mathbb N}$ and $n-k(n)$ members are from $(b\_n)\_{n\in\mathbb N}$.

**My Question:** When does the series $\sum\_{l=0}^{\infty} c\_l$ converges? Is there a good convergence test for this problem?

**My Attempt:** The expectational value of $c\_n$ should be

$$E(c\_n) = \frac{k(n)}{n} a\_n + \frac{n-k(n)}{n} b\_n$$

So I think, that $\sum\_{l=0}^{\infty} c\_l$ converges iff $\sum\_{l=0}^{\infty} E(c\_l)$ converges. Is this right? Can you point me to a theorem proving this?

Answer: 
Consider the sequence of functions on $(0,1)$
$$f\_n(x) = \begin{cases}
n & \text{ if } x \in (0,1/n)\\
0 & \text{ otherwise}
\end{cases}$$
We have $\lim\_{n \to \infty} f\_n(x) = 0 = f(x)$. However,
$$\lim\_{n \to \infty} \int\_0^1f\_n(x)dx = 1 \neq 0 = \int\_0^1 f(x) dx$$
The key in the dominated convergence theorem is that sequence of functions $f\_n(x)$ must be dominated by a function $g(x)$, which is also integrable.
Let $f\_n = \chi[n,2n]$ then for any $x \in \mathbb R$

$$\lim\_{n \to \infty} f\_n (x) = 0 \,\,\, \text{and} \,\,\, \int\_{x\in \mathbb R} f\_n(x) dx = n \to \infty \,\,\, \text{as}\,\, n \to \infty$$

On the other hand taking $f(x) = 0 , \forall x \in \mathbb R$ we have $$\int\_{x\in \mathbb R} f(x) dx = \int \_{x\in \mathbb R} 0\,\, dx = 0$$ then 

$$\int\_{x\in \mathbb R} f\_n(x) dx \not \to \int\_{x\in \mathbb R} f(x) dx$$