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Let $f(x)=\frac{1}{x+5}$

  1. Calculate $f^{(n)}(1)$ for $n \in \mathbb{N}$
  2. Find the sum $\sum_{n\geq 0} (-1)^n \frac{n4^{n-1}}{6^{n+1}}$

For the first item I find that $f^{(n)}(1)=(-1)^n\frac{n!}{6^{n+1}}$. I'm trying to use taylor series for the second item so I have that $f(x)=\sum_{n=0}^\infty (-1)^n \frac{(x-1)^n}{6^{n+1}}$ but I don't know how to use this for find $\sum_{n\geq 0} (-1)^n \frac{n4^{n-1}}{6^{n+1}}$

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Hint: Differentiate the Taylor series you wrote down term-by-term. Is there an $x$ for which this produces the desired series?

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We have

$$f(x)=\sum_{n=0}^{+\infty}\frac{(x-1)^n}{n!}f^{(n)}(1)$$

$$=\sum_{n=0}^{+\infty}(-1)^n\frac{(x-1)^n}{6^{n+1}}$$

and

$$f'(x)=\sum_{n=0}^{+\infty}(-1)^n\frac{n(x-1)^{n-1}}{6^{n+1}}$$

with $x=5$ and $f'(x)=-\frac{1}{(x+5)^2}$,

Your sum is

$$f'(5)=-\frac{1}{100}$$

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