So the power series \(\displaystyle \sum_{n=0}^{\infty}a_n (x-4)^n\) converges at 0 and diverges at 9. How do we know that the limit of the sequence \(\displaystyle a_n\) is zero? Thanks.
@tkhunny so I guess that would be"converges at 0" What do the terms look like for x = 0?
So the sequence would look like"converges at 0" What do the terms look like for x = 0?
So the sequence would look like
\(\displaystyle a_0+a_1(-4)+a_2(16)+a_3(-64)+...\)
I still don't see how the limit of \(\displaystyle a_n\), \(\displaystyle \lim_\limits{x\to\infty}a_n\), is 0?
@tkhunny so I guess that would be
\(\displaystyle a_0+a_1(-4)+a_2(16)+a_3(-64)...\)
I still don't see how the limit of the sequence \(\displaystyle a_n\), \(\displaystyle \lim\limits_{n\to\infty}a_n\) is 0?
Actually, having re-read that link to Paul's Online Maths Notes, it is indeed that a convergent series has a sequence that tends to 0. I can see the solution to the original problem now. However, isn't the power series, rather than the sequence $a_n$, that converges, and so wouldn't the theory only apply to the power series, and so we still can't conclude anything about \(\displaystyle a_n\) itself?