<s>[INST] Convergence of a series implies convergence of another series Let $a_1,a_2,\cdots$ be a sequence of real numbers with $a_i\geq 0$. If $\sum_{n=1}^{\infty}\frac{1}{1+a_n}<\infty$ then show that $\sum_{n=1}^{\infty}\frac{1}{1+x_na_n}<\infty$ for each real sequence $x_i$ with $x_i\geq 0$ and $\lim \inf x_n >0$.
Suppose $x_n\geq 1$ for all $n$ then $1+x_na_n\geq 1+a_n$ for all $n$. So, $\frac{1}{1+x_na_n}\leq \frac{1}{1+a_n}$ and in particular, we have 
$$\sum_{n=1}^{\infty}\frac{1}{1+x_na_n}<\sum_{n=1}^{\infty}\frac{1}{1+a_n}<\infty.$$
So, we are done in this case. 
We can assume that $\lim \inf x_n$ is less than $1$.
Just to get some idea, considering some simple sequences. Let $x_n=\frac{1}{2}+\frac{1}{n}$. Then $$\sum_{n=1}^{\infty}\frac{1}{1+x_na_n}=\sum_{n=1}^{\infty}\frac{2n}{2n+(n+2)a_n}=2\sum_{n=1}^{\infty}\frac{n}{2n+(n+2)a_n}=2\sum_{n=1}^{\infty}\frac{1}{2+(1+2/n)a_n}$$ 
Now, clealy $2+(1+2/n)a_n\geq 1+a_n$ which implies  $$\sum_{n=1}^{\infty}\frac{1}{1+x_na_n}=2\sum_{n=1}^{\infty}\frac{1}{2+(1+2/n)a_n}\leq 2\sum_{n=1}^{\infty}\frac{1}{1+a_n}<\infty.$$
So, we are done. There is nothing special about $2$.. Any $k$ would guarantee the convergence.
For $x_n=\frac{1}{k}+\frac{1}{n}$ then  $$\sum_{n=1}^{\infty}\frac{1}{1+x_na_n}=k\sum_{n=1}^{\infty}\frac{1}{k+(1+k/n)a_n}\leq k\sum_{n=1}^{\infty}\frac{1}{1+a_n}<\infty.$$
For this sequence, $x_n=\frac{1}{k}+\frac{1}{n}$, the $\lim \inf x_n $ is $\frac{1}{k}>0$. I want to do the same for general $1>\epsilon >0$. Given $1>\epsilon >0$ there exists $k$ such that $\epsilon>\frac{1}{k}$. 
So, $$\lim \inf x_n=\epsilon >\frac{1}{k}=\lim \inf \frac{1}{k}+\frac{1}{n}$$
Then i want to say $x_n\geq \frac{1}{k}+\frac{1}{n}$. So, $$\frac{1}{1+x_na_n}\leq \frac{1}{1+(1/k+1/n)a_n}$$ and in particular, 
$$\sum_{n=1}^{\infty}\frac{1}{1+x_na_n}\leq \sum_{n=1}^{\infty} \frac{1}{1+(1/k+1/n)a_n}\leq k\sum_{n=1}^{\infty}\frac{1}{1+a_n}<\infty.$$
I just want to know if there are any gaps or false statements.
 [/INST] Convergence only depends on the tail of the series. 
If $\liminf x_n \ge b > 0$, then there is some $N$ such that for $n > N$, $x_n > b/2$.  Let $B = \max(1, 2/b)$.
Then for $n > N$, 
$$\dfrac{1}{1+x_n a_n} \le \dfrac{1}{1 + (b/2) a_n} \le \dfrac{B}{1+a_n}$$
By a Comparison test, $\sum_n 1/(1+x_n a_n)$ converges.
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