Product of two infinite sequences Let $p_i$ be reals in (0,1) such that $\sum_1^{\infty} p_i=\infty$ and $\sum_1^{\infty} (1-p_i)=\infty$. Prove that $\sum_1^{\infty} p_i(1-p_i)=\infty$. I know a probabilistic proof (follows from Kolmogorov 0-1 Law on infinite seq of independent Bernoulli random variables) but I am unable to show it by usual real analysis trick. Any kind of help/hint is highly appreciated. And this has just come to my mind out of the curiosity.
Edit: $\lim p_n$ exists.
 A: This is false, when $\lim_{n \to \infty} p_n$ doesn't exist. For instance,
consider $p_{2n-1} = \dfrac1{2^n}$ and $p_{2n} = 1-\dfrac1{2^n}$. This gives us $\sum_{n=1}^{2N} p_n = \sum_{n=1}^{2N} (1-p_n) = N$, whereas $$\sum_{n=1}^{2N} p_n(1-p_n) < \sum_{n=1}^N \dfrac1{2^{n-1}} < 2$$

However, the claim is true when $\lim_{n \to \infty} p_n$ exists.
When $\lim_{n \to \infty} p_n$ exists, we have $3$ options:


*

*If $\lim_{n \to \infty} p_n \in (0,1)$, then again $\lim_{n \to \infty}  p_n(1-p_n) \neq 0$. Hence, $\sum_n p_n(1-p_n)$ diverges.

*If $\lim_{n \to \infty} p_n = 0$, then we have $\lim_{n \to \infty} \dfrac{p_n(1-p_n)}{p_n} = 1$. Hence by limit comparison test, $\sum_n p_n(1-p_n)$ diverges since $\sum_n p_n$ diverges.

*If $\lim_{n \to \infty} p_n = 1$, then we have $\lim_{n \to \infty} \dfrac{p_n(1-p_n)}{1-p_n} = 1$. Hence by limit comparison test, $\sum_n p_n(1-p_n)$ diverges since $\sum_n (1-p_n)$ diverges.
A: This is false.
Let $$p_i=\begin{cases} \frac{1}{2^i}&i\text{ odd}\\1-\frac{1}{2^i}&i\text{ even}\end{cases}$$
Then $p_i(1-p_i)<\frac{1}{2^i}$, so $\sum p_i(1-p_i)$ converges.
A: It all occurs from the following problem,
"$X_n$'s are independent $Bernoulli(p_n)$ random variables with
$$ \sum_1^{\infty} p_i=\infty=\sum_1^{\infty} (1-p_i)$$
Then $P\Bigl\{[X_n=1,X_{n+1}=0]\textit{ i.o.}\Bigr\}=1$"
The proof goes along the following lines
$$\Bigl[[X_n=1]\textit{ i.o.}\Bigr]\cap \Bigl[[X_n=0]\textit{ i.o.}\Bigr]\subseteq\Bigl[[X_n=1,X_{n+1}=0]\textrm{ i.o.}\Bigr]$$
(In fact '=' holds in place of '$\subseteq$', but we only care about $\subseteq$). By hypothesis and Borel-Cantelli Lemma 2
$$ P\Bigl\{[X_n=1]\textit{ i.o.}\Bigr\}=1= P\Bigl\{[X_n=0]\textit{ i.o.}\Bigr\}$$
As $X_n$'s are independent we have $P\Bigl\{[X_n=1 \textit{ i.o.}]\cap[X_n=0 \textit{ i.o.}]\Bigr\}=1$ and hence the result follows from the set inequality.
Here are we assuming $\lim P(X_n=1)$ exists?
I was trying from a different angle. Define
$$A_n=\Bigl[X_{2n-1}=1,X_{2n}=0\Bigr] \quad n=1,2,3...$$
Then $A_n$'s are independent and $P(A_n)=p_{2n-1}(1-p_{2n})$, If we can show the following
"$p_i$ $\in$ $(0,1)$ such that $\sum_1^{\infty} p_{i}=\infty$ and $\sum_1^{\infty} (1-p_{i})=\infty$, then $\sum_1^{\infty} p_{2i-1}(1-p_{2i})=\infty$".
Then again by Borel-Cantelli Lemma $\sum P(A_n)=\sum_1^{\infty} p_{2n-1}(1-p_{2n})=\infty$ implying $P(A_n \textit{ i.o.})=1$. As a matter of fact being not able to prove the above I ended up posting a wrong (with probability 1) statement and I do apologize for wasting everyone's time.
