binomial distributionnormal distributionpoisson distributionprobability distributions
I'm reading two books and they say differently.
In a binomial distribution $X \sim \text{Bin}(n,p)$, if $n \to +\infty$, $X$ approaches to Poisson distribution $\text{Po}(np)$.
The other book says $X$ approaches to normal distribution $\text{N}(np,np(1-p))$.
I'm confused.
In comments above, Tim asked why it must be that if $X\sim\mathrm{Poisson}(\lambda)$ and $Y\sim\mathrm{Poisson}(\mu)$ and $X$ and $Y$ are independent, then we must have $X+Y\sim\mathrm{Poisson}(\lambda+\mu)$.
Here's one way to show that. \begin{align} & \Pr(X+Y= w) \\[8pt] = {} & \Pr\Big( (X=0\ \& \ Y=w)\text{ or }(X=1\ \&\ Y=w-1) \\ & {}\qquad\qquad\text{ or } (X=2\ \&\ Y=w-2)\text{ or } \ldots \text{ or }(X=0\ \&\ Y=w)\Big) \\[8pt] = {} & \sum_{u=0}^w \Pr(X=u)\Pr(Y=w-u)\qquad(\text{independence was used here}) \\[8pt] = {} & \sum_{u=0}^w \frac{\lambda^u e^{-\lambda}}{u!} \cdot \frac{\mu^{w-u} e^{-\mu}}{(w-u)!} \\[8pt] = {} & e^{-(\lambda+\mu)} \sum_{u=0}^w \frac{1}{u!(w-u)!} \mu^u\lambda^{w-u} \\[8pt] = {} & \frac{e^{-(\lambda+\mu)}}{w!} \sum_{u=0}^w \frac{w!}{u!(w-u)!} \mu^u\lambda^{w-u} \\[8pt] = {} & \frac{e^{-(\lambda+\mu)}}{w!} (\lambda+\mu)^w \end{align} and that is what was to be shown.
Okay, after some investigating, I have learned some things about statistics. I will post this answer here in the hope that someone finds it helpful in the future. Thanks to helpful comments by @spaceisdarkgreen.
Essentially what this boils down to is the Central Limit Theorem (Wikipedia). The ``usual'' CLT applies to the sum of identically distributed random variables--that is, variables drawn from the same distribution. That does not apply in the case of the Poisson-Binomial distribution, since each variable in the sum is drawn from a Bernoulli distribution with a different mean. Thus, we need a generalization of the CLT for non-identically distributed random variables. Of course, this will require some additional assumptions on the variables, but fortunately they are easily satisfied by Bernoulli random variables.
The necessary modification is provided by the Lyapunov Central Limit Theorem (Wikipedia), (MathWorld) (note that Wikipedia uses $\mathbf{E}[\cdot]$ to denote expectation, while MathWorld uses $\langle\cdot\rangle$). Also, see this answer for a related discussion.
Anyway, the Poisson-Binomial distribution satisfies the Lyapunov condition, and hence, loosely speaking, the Poisson-Binomial distribution will converge to the normal distribution with mean and variance
$$\mu = \sum_{i=1}^n p_i, \quad \sigma^2 = \sum_{i=1}^n p_i(1-p_i)$$
respectively. To confirm this, I tested with means $p_i$ spaced uniformly between $p_0 = 0.35$ and $p_{n-1}=0.65$ for multiple values of $n$. Two plots are shown below (note that the probability mass function of the Poisson-Binomial distribution was computed via Monte Carlo sampling with $N=50\ 000%$ points, since computing the pmf explicity can become a little tricky. 
These results suggest that in practice, the convergence may be relatively quick, with reasonable agreement after only $n=6$ (NOTE that in the case where you wish to sum an infinite sequence of Bernoulli random variables, you would require that the $p_i$ be bounded away from 0 and 1! This didn't bother me because I am interested in a finite sequence).
Best Answer
Either the two books are sloppy or you're not reading precisely.
If $n\to\infty$ and $p\to0$ while $np$ approaches some positive number $\lambda,$ then the binomial distribution approaches a Poisson distribution with expected value $\lambda.$
If $n\to\infty$ as $p$ stays fixed, and $X\sim\operatorname{Binomial}(n,p)$ then the distribution of $(X-np)/\sqrt{np(1-p)}$ approaches the standard normal distribution, i.e. the normal distribution with expected value $0$ and standard deviation $1.$
It is sloppy to say something approaches something depending on $n$ as $n\to\infty,$ unless it is precisely defined and not meant literally. Thus the statement that something approaches $\operatorname{Binomial}(np, np(1-p))$ as $n\to\infty$ is not to be taken literally, but rather it means what is stated in the second bullet point above, where the limit, the standard normal distribution, does not depend on $n.$ In the first bullet point above, the statement that something approaches $\operatorname{Poisson}(np)$ can make sense only because $np$ does not depend on $n,$ that is, what is considered is a limit as $n\to\infty$ and $p\to0$ which $np$ remains fixed.