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The Taylor series expansion would be,\[\sum_{n=0}^{\infty}=\frac{(-1)^n}{2^n(n+2)}(x-5)^n\]

You need to expand this up to the third power; that is, from n=0 to 3.

I'm finding the radius of convergence.

Hey, that's good :) So are you right to do the radius of convergence?

Ahaha...unfortunately no. Mind giving me a hint? xD

Okay. Just give me a minute to get a drink.

\[=\left| x-5 \right|\frac{1+2/n}{2+6/n}\]

We need to find all x such that the above holds true.

So we have three regions to test:
\[-\infty < x< 3\]\[3

Your radius of convergence is\[\left\{ x \in \mathbb{R} |3

Is that what you mean?

No worries - just fan me (if you haven't already)! ;)