$\sum_1^{\infty}\frac1{1+x_na_n}$ converges if $\sum_1^{\infty}\frac1{1+a_n}$ converges Let $\langle x_n\rangle$ and $\langle a_n\rangle$ be two positive sequences of real numbers such that $\displaystyle\liminf_{n\to\infty}x_n\gt0$ and $\displaystyle\sum_1^{\infty}\frac1{1+a_n}<\infty$. Then, can we conclude that $\displaystyle\sum_1^{\infty}\frac1{1+x_na_n}<\infty$?
I think yes. But, my guess is intuitive. I think since $\displaystyle\liminf_{n\to\infty}>0$, we can bound the inverse sum $\displaystyle\sum_1^{\infty}\frac1{1+x_n}$ and then use the Cauchy Schwarz inequality to get the result. Any ideas. Thanks beforehand.
 A: The head of the sequence makes no difference, so let $m=\liminf x_n$ and $\epsilon$ with $m>\epsilon >0$ be given and let $N>\!\!\!>0$ be so that for all $n\ge N$ $x_n>m-\epsilon$. Then we can see by direct comparison that

$$\sum{1\over 1+a_nx_n}<\sum {1\over 1+a_n(m-\epsilon)}$$

Now as $\sum{1\over 1+a_n}$ converges $a_n\to\infty$ so by limit comparison test, the RHS converges and by normal comparison test the desired series converges.
A: Let's take $N, \delta$ such that $(\forall n \geq N)\delta < x_n$ and $0 < \delta < 1$. Then $$\sum_{n \geq N} \frac{1}{1 + x_n a_n} < \sum_{n\geq N}\frac{1}{1 + \delta a_n} = \frac{\delta^{-1}}{\delta^{-1} + a_n} = \delta^{-1} \sum_{n \geq N} \frac{1}{\delta^{-1} + a_n} < \delta^{-1} \sum_{n \geq N} \frac{1}{1 + a_n} < \infty$$
(the second inequality comes from the fact that $\delta^{-1} > 1$, so the denominator of each term in the left-hand sum exceeds the denominator of the corresponding term in the right-hand sum). And of course the terms for $n < N$ can't affect convergence, so $\sum_{n \geq 1} \frac{1}{1 + x_n a_n}$ converges.
