Inequality for probability distribution Suppose that $\sum_{I=1}^{\infty}p_i=1$ and $p_i\geq 0$. Show that $\sum_{i}p_i(1-p_i)^n$ goes to $0$ as $n \rightarrow \infty$.
My idea: I tried to use the inequality $(1-p_i)^n\geq 1-np_i$ but I think we need to use an inequality in opposite side.
 A: What you should notice is that each term converges to zero: $\lim_{n\to\infty} p_i(1-p_i)^n = 0$. So the question is, how to harness this observation to produce the same conclusion for the infinite sum $\sum_{i=1}^{\infty} p_i(1-p_i)^n$. A natural idea will be to interchange the order of limit and summation. However, we also know that this is not always possible, so you will end up seeking a mathematical result that ensures this.
As many users pointed out, either monotone convergence theorem or dominated convergence theorem fits in your situation. Although not mentioned in comments, even Weierstrass $M$-test can be used to justify this. And the common reason why all these fancy techniques are available is that we have a summable dominating sequence $\{p_i\}$ for $\{p_i(1-p_i)^n\}$.
As a final blow, let me provide a low-tech solution (which of course critically rely on the above observation) : Fix $N$ and take limsup as $n\to\infty$ to
$$ \sum_{i=1}^{\infty} p_i (1-p_i)^n \leq \left( \sum_{i=1}^{N} p_i (1-p_i)^n \right) + \left( \sum_{i=N+1}^{\infty} p_i \right). $$
Then using the subadditivity of limsup, we have
$$ \limsup_{n\to\infty} \sum_{i=1}^{\infty} p_i (1-p_i)^n \leq \sum_{i=N+1}^{\infty} p_i. $$
Since the limsup in the left is a fixed number which is independent of $N$, we can let $N\to\infty$ to both sided to obtain $\limsup_{n\to\infty} \sum_{i=1}^{\infty} p_i (1-p_i)^n \leq 0$. Since the complementary inequality for liminf is trivial, it follows that the sum converges to $0$ a $n\to\infty$.
A: Been a while, but I think this does it. 
Let a = $\sum_{i}p_i(1-p_i)^n$ and b=$\sum_{i}(1-p_i)^n$. 
and c= $\sum_{i}(1-p_i)$^(n+1).
b-a=c, so b=a+c.
$0<p_i$<1, so $p_i^k$<$p_i^m$ if k>m. We also have $p_i(1-p_i)^n$<$(1-p_i)^n$ since (1-$p_i$ ) is also positive.
Note c and b have the same necessarily finite limit, so we avoid indeterminate terms. a being equal to b-c, with c and b being finite and determinate, lim a = lim (b-c)=0.
