# Prove $\{a_n\}$ converges to zero with $a_{n+1}\leqslant (1-\lambda_n)a_n+b_n+c_n$

Let $a_n$,$b_n$,$c_n$be sequences of nonnegative numbers satisfying ,for all integers $n\geq 0$, \begin{align*} a_{n+1}\leqslant (1-\lambda_n)a_n+b_n+c_n \end{align*} with $\lambda_n\in [0,1]$,$\sum_{n=0}^{+\infty}\lambda_n=+\infty$,$b_n=o(\lambda_n)$as$n\to \infty$,and$\sum_{n=0}^{+\infty}c_n<+\infty$.Prove that the sequence $\{a_n\}$ converges to zero.

Once we have proved that $\{a_n\}$ converges,then we can conclude that $\lim_{n\to \infty}a_n=0$.In fact,we set $\lim_{n\to \infty}a_n=\alpha$,then $\alpha\geq 0$.Suppose $\alpha>0$,then we can find an integer $N$,such that for all $n\geq N$, \begin{align*} a_{n+1}\leqslant &a_n-\lambda_n(a_n-\frac{b_n}{\lambda_n})+c_n\\ <&a_n-\frac{\alpha}{2}\lambda_n+c_n\\ \Rightarrow \frac{\alpha}{2}\lambda_n<&a_n-a_{n+1}+c_n. \end{align*}

then we can get \begin{align*} \frac{\alpha}{2}\sum_{n=N}^\infty\lambda_n<a_N-\alpha+\sum_{n=N}^\infty c_n<+\infty. \end{align*} which is a contradiction to the fact that $\sum_{n=0}^{+\infty}\lambda_n=+\infty$!So it must be that $\alpha=0$.

What puzzles me is how to prove $\{a_n\}$ converges.Can anyone give me some help?Thanks in advance.

I have got it！We should discuz the relation between$a_n$and$d_n=：\frac{b_n}{\lambda_n}$。One case is there are finite numbers of $a_n$which is not greater than $d_n$，then when $n$is large enough ,we have$a_{n+1}\leqslant a_n+c_n$，and by this We can prove that$\{a_n\}$is convergent，and associating with the preceding analysis，It must be convergent to 0；The other case,there are infinite numbers of $a_n$which is not greater than $d_n$,We pick them out and can construct a sequence which is convergent to zero,then We can easily prove that $\{a_n\}$converges to 0.

## 1 Answer

Update: The previous answer (see below in "wrong answer") was a wrong proof. Thanks to Chen Jiang for an example. Series convergence is not necessary as seen in the example.

So a new approach towards solving (hopefully) this question: Observe that $$a_{n+1} - a_n \leq b_n + c_n - \lambda_n a_n.$$ Rewrite as $$a_{n+1} - a_n \leq \lambda_n \frac{b_n}{\lambda_n} + c_n - \lambda_n a_n,$$ whenever $\lambda_n \neq 0$.

Now since $\lambda_n \in [0,1]$, rewrite the above inequality as, $$a_{n+1} - a_n \leq \frac{b_n}{\lambda_n} + c_n.$$ We know that $c_n \rightarrow 0$ and $\frac{b_n}{\lambda_n} \rightarrow 0$, thus $a_{n+1} - a_n < \epsilon$ for sufficiently large $n$.

Claim: $a_{n} - a_{n+1} < \epsilon$ for sufficiently large $n$.

If this claim is true then the limit of $\{a_n\}$ exists. QED.

So any hints on how to prove the above claim? If one is unsuccessful, try to get a counter example for the question, which I am not getting until now.

Wrong answer: Claim: $\sum a_n$ is a convergent series.

We know that If a series converges, then the sequence of terms converges to $0$. $\blacksquare$

Hint for proof of the claim, use the ratio test (i.e., https://en.wikipedia.org/wiki/Ratio_test#The_test).

• I'm afraid your claim could not be true,either .For example,take $a_n=\sin^2\sqrt{n}$,then $|a_{n+1}-a_n|=\frac{\sin \sqrt{\xi_n}}{\sqrt{\xi_n}}\to 0$,while $\{a_n\}$itself is not convergent. – mbfkk Feb 12 '17 at 2:24