Sum of quantities of infinite number of coin flip probabilities Suppose there are $N$ coins with head probabilities being $P_{1}$ through $P_N$. We flip them altogether. Let's use $Q(l), l=0,1,\dots,N$ to denote the prob. of exact $l$ out of these $N$ coins turn head after flip. Obviously we have $Q(N)=\prod_{i=1}^{N}P_{i}$ and so on.
What we are interested is the quantity of $S \triangleq\sum_{l=0}^N\frac{Q(l)}{l+1}$.
As a special example, when $P_{1}=P_{2}=...=P_{N}=0.5$, we can easily get
$S=\sum_{l=0}^N\frac{Q(l)}{l+1}=\frac{2^{N+1}-1}{(N+1)2^N}$. Thus it is safe to say that $S$ is $O(\frac{1}{N})$ as $N$ goes to $+\infty$.
But this is too restrictive. I would like to know if there is a general enough condition on $P_{1}$ thought $P_{N}$ so that I can show that $S$ is $O(\frac{1}{N})$.
 A: I'll write $N$ as $n$ and $\ell$ as $k$. We have a generating function
$$f(x) = \sum_{k=0}^n Q(k) x^k = \prod_{i=1}^n ((1 - p_i) + p_i x)$$
which gives
$$\sum_{k=0}^n \frac{Q(k)}{k+1} = \int_0^1 f(x) \, dx = \int_0^1 \prod_{i=1}^n ((1 - p_i) + p_i x) \, dx.$$
So now we need to bound this integral. The easiest special case is the one where each of the $p_i$ is equal to some $p$, where the integrand becomes $f(x) = ((1 - p) + px)^n$, which gives
$$S(p) = \int_0^1 ((1 - p) + px)^n \, dx = \frac{1 - (1 - p)^{n+1}}{p(n+1)}.$$
So we see that the integral is $O \left( \frac{1}{pn} \right)$ in this case. This suggests we'll run into the issues if the probabilities $p_i$ get too small. If $p_i \ge p > 0$ for some fixed $p$ then we have $(1 - p_i) + p_i x \le (1 - p) + px$ so the integral is bounded above by $S(p)$ and then we're fine. But if $p_i \to 0$ then we're in trouble; are you interested in this case?
A: I conjecture that it's enough that for some $0<a<1/2$ we have $a\le p_i < 1-a$ (i.e. not too close to $0$ or $1$).
Calling $X$ the amount of heads, $\mu = \sum p_i$ and $\sigma^2=\sum p_i (1-p_i)$. If both grow as $N$, and the condition above apply, then we can assume CLT and do a Taylor expansion , so that
$$E\left[\frac{1}{X+1}\right] \approx \frac{1}{\mu+1} - \frac{\sigma^2}{(\mu+1)^3} \sim \frac{1}{N}$$
