# Thread: convergence of a sequence

1. ## 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.

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

3. 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.

4. Let $\displaystyle \displaystyle t_n:=b\sum_{j=0}^nx^j$. If we denote by $\displaystyle d_n:=s_n-t_n$ we get that $\displaystyle d_{n+1}=xd_n+b_{n+1}-b$ hence $\displaystyle |d_{n+1}|\leq|x||d_n|+|b_{n+1}-b|$. Let $\displaystyle l:=\limsup_{n}|d_n|$; we have $\displaystyle l\leq |x|l$ hence $\displaystyle l=0$.

5. 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)

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...

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

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

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

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

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
We know that the sequence $\displaystyle \left\{b_n\right\}$ is convergent (to $\displaystyle b$) but not the series.

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