Prove that this bounded sequence is convergent

Let $(a_n)_{n \geq 1} \subset \mathbb{R}$ and $|q| \in (0,1)$. Define $$x_n=\sum_{k=1}^na_kq^k$$ Prove that if $(a_n)_{n \geq 1}$ is bounded, then $(x_n)_{n \geq 1}$ is convergent. Also, give an example of a sequence $(a_n)_{n \geq 1}$, unbounded, such as $(x_n)_{n \geq 1}$ to be convergent.

For the first part, I proved that $(x_n)_{n \geq 1}$ is bounded, since there is a positive $M$ such that $|a_n| \leq M, \forall n$: $$|x_n|=|\sum_{k=1}^na_kq^k| \leq \sum_{k=1}^n|a_k||q^k| \leq M \sum_{k=1}^n|q^k|=M|q|\frac{1-|q|^n}{1-|q|} \leq M\frac{|q|}{1-|q|}$$ Obviously, since $a_n$ can be either positive or negative, $(x_n)_{n \geq 1}$ is not necessary monotonic and thus the convergence doesn't follow immediately.

• To show that $(x_n)_n$ is a Cauchy sequence use the same method to get an upper bound $B(m+n,n)$ for $|x_{m+n}-x_n|$. – DanielWainfleet Sep 18 '17 at 21:59

$$\sum_{k=1}^{\infty}|a_kq^k| \leq M \frac{q}{1-q}$$

using the fact that $|a_k| \leq M,\forall k \in \mathbb{N}$

Thus the series converges absolutely,hence converges to some $a \in \mathbb{R}$

so $\lim_{n \to + \infty}{x_n}=a$ which means that the limit of $x_n$ exists.

For the second part of your question take $a_k=k$ and $q=\frac{1}{2}$

In fact any $q \in (0,1)$ will do.

You can use the ratio test for $b_k=\frac{k}{2^k}$(or for $b_k=kq^k$) to see that $\sum b_k$ converges

proving that $x_n=\lim_{n \to +\infty}\sum_{k=1}^nb_n$ has a limit.