# convergence of a sequence

• Mar 24th 2011, 09:40 AM
convergence of a sequence
Hi,

I have a problem in my mind, it is as follows:

We know that the sum of positive integer powers of a number x s.t. x is between 0 and 1 converges to 1/(1-x). Furthermore, assume that we have sequence b_{n} converging to b>0. Let

s_{n} = {sum}_{i=0}^{n}[x^{n-i}*b_{i}] (in words: s_{n} equals to sum from i=0 to n of the terms x to the n-i multiplied by b_{i})

The question is, does s_{n} converges?{may be helpful: if it converges, it converges to b/(1-x)}.

Thanks.
• Mar 24th 2011, 09:57 AM
girdav
We can show that $s_{n+1}=xs_n+b_{n+1}$ by computing $s_{n+1}-s_n$.
• Mar 24th 2011, 10:04 AM
Yes, i actually did that, then i think the following can be asserted:

s_{n} converges if and only if s_{n+1} - s_{n} = (x-1)*s_{n} + b_{n+1} converges to 0. I run many MATLAB codes those implying the sequence indeed converges but i couldn't show it rigoruously.
• Mar 24th 2011, 10:44 AM
girdav
Let $\displaystyle t_n:=b\sum_{j=0}^nx^j$. If we denote by $d_n:=s_n-t_n$ we get that $d_{n+1}=xd_n+b_{n+1}-b$ hence $|d_{n+1}|\leq|x||d_n|+|b_{n+1}-b|$. Let $l:=\limsup_{n}|d_n|$; we have $l\leq |x|l$ hence $l=0$.
• Mar 24th 2011, 01:43 PM
It's very nice actually, do you think we need to add that limsup(d_n)=liminf(d_n)=0? (since limit exists iff limsup=liminf)
• Mar 24th 2011, 01:51 PM
chisigma
Quote:

Hi,

I have a problem in my mind, it is as follows:

We know that the sum of positive integer powers of a number x s.t. x is between 0 and 1 converges to 1/(1-x). Furthermore, assume that we have sequence b_{n} converging to b>0. Let

s_{n} = {sum}_{i=0}^{n}[x^{n-i}*b_{i}] (in words: s_{n} equals to sum from i=0 to n of the terms x to the n-i multiplied by b_{i})

The question is, does s_{n} converges?{may be helpful: if it converges, it converges to b/(1-x)}.

Thanks.

Applying the Cauchy rule for the product of two converging series You have...

$\displaystyle \sum_{n=0}^{\infty} \sum_{i=0}^{n} x^{n-i}\ b_{i} = \sum_{n=0}^{\infty} x^{n}\ \sum_{i=0}^{\infty} b_{i} = \frac{b}{1-x}$

Kind regards

$\chi$ $\sigma$
• Mar 24th 2011, 02:00 PM
girdav
Quote:

Originally Posted by chisigma
Applying the Cauchy rule for the product of two converging series You have...

$\displaystyle \sum_{n=0}^{\infty} \sum_{i=0}^{n} x^{n-i}\ b_{i} = \sum_{n=0}^{\infty} x^{n}\ \sum_{i=0}^{\infty} b_{i} = \frac{b}{1-x}$

Kind regards

$\chi$ $\sigma$

We know that the sequence $\left\{b_n\right\}$ is convergent (to $b$) but not the series.
• Mar 24th 2011, 02:04 PM
girdav
Quote:

Yes it's a step of our proof, even if someone who read until " $l=0$" understands that the limit exists. We have to add that $\left\{t_n\right\}$ is convergent and $\left\{s_n\right\}$ is also convergent and has the same limit.