∑[n=1,∞,(15-x^(2))^(n)]

Question

anonymous · Answer

well here is the answer, lets see if we can get there http://www.wolframalpha.com/input/?i=sum+n+%3D+1+to+infinity+%2815-x^2%29^n

jamesj · Answer

Remember first that for |r| < 1,
$$ 1 + r + r^2 + ... + r^n + r^{n+1} + ... = \frac{1}{1-r} $$

Now use that relation with your expression above.  I.e., r = 15 - x^2

anonymous · Answer

think it might be a simple as using 
$$\sum x^n=\frac{1}{1-x}$$ with some adjustments

anonymous · Answer

what jamesj said

jamesj · Answer

Deriving the formula for the sum is pretty straight-forward.  What requires a few more neurons is determining for what values of x the infinite sum exists.

anonymous · Answer

well i have a question. why the "-1" i probably would have missed that one

jamesj · Answer

because in the 'geometric series' of the question, there is no "1" term in the sum.  It starts with r; not 1.

anonymous · Answer

right. i often mess that one up, unless i start from scratch.  as for radius of convergence, it is pretty damned small i think

jamesj · Answer

It's the x such that
$$ |15-x^2| < 1 $$
It's a more interesting set than most convergence intervals.

jamesj · Answer

@charitha, you following?

anonymous · Answer

@JamesJ yes,,
thanks
:)

anonymous · Answer

oh there are two intervals! is this part of the question?

anonymous · Answer

@satellite73 
ya