Question

Proof Question coming

Answer 1

And by convergence tests: ratio, root, limit comparison, any other.

Answer 2

Use the ratio test. We need:
\[\lim_{k \rightarrow \infty}a_k \rightarrow 0; \left| \lim_{k \rightarrow \infty} \frac{a_{k+1}}{a_k}\right| <1\]
So:
\[\left| \lim_{k \rightarrow \infty} \frac{(k+1)^2(x+3)^k (x+3)}{k^2 (x+3)^k}\right|=\left| \lim_{k \rightarrow \infty} \left( \frac{k+1}{k}\right)^2 (x+3)\right|=\left| x+3 \right|<1\]
So:
\[-1 < x+3 <1 \implies -4 < x <-2\]
Plug in -4 for x to check the endpoint on the left:
So:
\[\sum_{k=0}^{\infty}k^2 (-1)^k\]
This is divergent so you do NOT include the left endpoint. Now check the right, -2.
\[\sum_{k =0}^{\infty}k^2*(1)^k\]
This is also divergent. So the interval of convergence is:
\[x \in (-4,-2)\]

Answer 3

Do you follow that?

Answer 4

yea

Answer 5

do you think you could help on one more?

Answer 6

Sure