2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ To show that $(x_n)_n$ is a Cauchy sequence use the same method to get an upper bound $B(m+n,n)$ for $|x_{m+n}-x_n|$. $\endgroup$ – DanielWainfleet Sep 18 '17 at 21:59
1
$\begingroup$

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

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.