Real Analysis

[math25]

Hi,

Can someone please help me with this problem

Let {ak) be a series with members 0 (lessthan or equal t)o ak ( lessthen or equal to) 1

Show that the series sum akx^k converges for all values 0 l(essthan or equal to) x (less then 1)

Thanks

 

[daon2]

\(\displaystyle x^k a_k \le (x^k)(1) = x^k\)

If necessary, you may use that the partial sums \(\displaystyle S_n = \sum_{k=1}^n a_kx^k\) is increasing and bounded above (by what?).

 

[math25]

its bounded by 1, right?

 

[pka]

its bounded by 1, right?

More importantly, we know that
\(\displaystyle \displaystyle \sum\limits_{k = 1}^\infty {a_k x^k } \le \sum\limits_{k = 1}^\infty {x^k } = \frac{{x }}{{1 - x}}\)

 

[math25]

1/(1-x)= 1 + x + x^2 +...

Sn= sum ak x^k

since series are convergent for any 0 less than or equal x less than 1 we have :

Sn / (1-x) = sum (a0 + a1 + a2 +...ak) x^k

sk=(a0 + a1 + a2 +....ak) = sum ak

and I am not sure what to do from here...

 

[pka]

I am not sure what to do from here...

By the basic comparison test the given series converges.