Taylor Series — Question 8

Question 8 – Solution

Step 1: Compute Taylor coefficients.

Repeated differentiation gives $$f^{(n)}(x)=\frac{n!}{(1-x)^{n+1}},\qquad f^{(n)}(0)=n!.$$ Thus the Maclaurin coefficient is $f^{(n)}(0)/n!=1$, and $$\boxed{\frac1{1-x}=\sum_{n=0}^{\infty}x^n}.$$

Step 2: Verify using finite sums.

$$S_N=1+x+\cdots+x^N=\frac{1-x^{N+1}}{1-x}.$$ If $|x|<1$, then $x^{N+1}\to0$, so $S_N\to1/(1-x)$. This proves the Taylor series and geometric identity are the same construction.

Step 3: Check endpoints.

At $x=1$ the terms are $1$; at $x=-1$ they alternate between $\pm1$. Both diverge by the nth-term test.

Conclusion.

$R=1$ and the interval is $\boxed{(-1,1)}$.