convergence of a sequence

**MADonmez** · March 24th 2011, 09:40 AM

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.

**girdav** · March 24th 2011, 09:57 AM

We can show that $s_{n+1}=xs_n+b_{n+1}$ by computing $s_{n+1}-s_n$ .

**MADonmez** · March 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.

**girdav** · March 24th 2011, 10:44 AM

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 $l=0$ .

**MADonmez** · March 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)

**chisigma** · March 24th 2011, 01:51 PM

Originally Posted by MADonmez

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

$\chi$ $\sigma$

**girdav** · March 24th 2011, 02:00 PM

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.

**girdav** · March 24th 2011, 02:04 PM

Originally Posted by MADonmez

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)

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.

**MADonmez** · March 24th 2011, 02:53 PM

thanks, but b_{n} is a "sequence" converging to b, so the series above diverges

**MADonmez** · March 24th 2011, 03:11 PM

thank you very much

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