I am puzzled by this. The "Taylor series for f(x) at x= a" is pretty well defined in any Calculus text:
\(\displaystyle \sum_{n=0}^\infty \frac{1}{n!}\frac{d^nf}{dx^n}(x- a)^n\).
Typically the hardest part is to find the general \(\displaystyle \frac{d^nf}{dx^n}\). So just start taking derivatives and see if you can see a pattern. Here \(\displaystyle f(x)= \frac{4}{x+1}=4(x+1)^{-1}\) and a= 2.
\(\displaystyle f(2)= \frac{4}{3}\).
\(\displaystyle f'(x)= -4(x+ 1)^{-2}\) so \(\displaystyle f'(2)= -\frac{4}{3^2}= -\frac{4}{9}\).
\(\displaystyle f''(x)= 8(x+1)^{-3}\) so \(\displaystyle f''(2)= \frac{8}{3^3}= \frac{8}{27}\)
\(\displaystyle f'''(x)= -24(x+ 1)^{-4}\) so \(\displaystyle f'''(2)= -\frac{24}{3^4}= -\frac{24}{81}\)
\(\displaystyle f''''(x)= 96(x+ 1)^{-5}\) so \(\displaystyle f''''(2)= \frac{96}{3^5}= \frac{96}{243}\)
\(\displaystyle f''''''(x)= -480(x+ 1)^{-6}\) so \(\displaystyle f'''''(2)= -\frac{480}{3^6}= -\frac{480}{729}\)
Can you see a pattern yet? I see an alternating sign. Each even n term is positive while each odd n term is negative: that's \(\displaystyle (-1)^n\). The denominators are powers of 3: that is \(\displaystyle 3^n\). The numerators are 4, 8, 24, 96, 480 .... That's a little harder! Of course, each number is divisible by 4: they are 4 times 1, 2, 6, 24, 120, ....
Aha! The numbers 1, 2, 6, 24, 120 are 1!, 2!, 3!, 4!, and 5! I will conjecture that the numerators are 4n! which means that the coefficent of \(\displaystyle (x- a)^n\) is \(\displaystyle (-1)^n\frac{1}{n!}\frac{4n!}{3^n}\)
and the Taylors series is \(\displaystyle \sum_{n=0}^\infty (-1)^n\frac{1}{n!}\frac{4n!}{3^n}(x- 2)^n\)\(\displaystyle = \sum_{n=0}^\infty (-1)^n\frac{4}{3^n}(x-2)^n\).
Of course, I will need to prove the conjecture that nth numerator of the derivative is 4n!. That can be done by "induction".
It is pretty standard to determine the radius of convergence using the "ratio test" (sometimes the "power test" is used). An infinite series, \(\displaystyle \sum a_n\), converges absolutely if \(\displaystyle \lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right|< 1\). For a power series, \(\displaystyle sum a_n (x-a)^n\), that becomes \(\displaystyle \lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right||x-a|\).
Here, the nth term is \(\displaystyle (-1)^n\frac{4}{3^n}\) so that \(\displaystyle \left|\frac{a_{n+1}}{a_n}\right||x-a|\) is \(\displaystyle \frac{4}{3^{n+1}}\frac{3^n}{4}|x- 2|= \frac{1}{3}|x-2|< 1\) so \(\displaystyle |x- 2|< 3\).
