<s>[INST] power series representation of a function by differentiation Find power series representation for $\frac{1}{(7+x)^2}$
What I tried...
$$\frac{1}{1-x}=\sum_{n=0}^{\infty}x^n$$
$$\frac{1}{7(1-(-{x \over7}))}=\sum_{n=0}^{\infty}\frac{(-1)^nx^n}{7^{n+1}}$$
$${d\over dx}(\frac{1}{7+x})={d\over dx}(\sum_{n=0}^{\infty}\frac{(-1)^nx^n}{7^{n+1}}$$
$$\frac{1}{(7+x)^2}=\sum_{n=1}^{\infty}\frac{(-1)^nnx^{n-1}}{7^{n+1}}$$
$$\frac{1}{(7+x)^2}=\sum_{n=0}^{\infty}\frac{(-1)^{n+1}(n+1)x^n}{7^{n+2}}$$
The correct answer:$$\sum_{n=0}^{\infty}\frac{(-1)^{n}(n+1)x^n}{7^{n+2}}$$
I don't understand why we are not adding +1 to the $(-1)^n$ term when we reduce n from 1 to 0.
 [/INST] Hint: Instead of differentiation you could also apply the formula for the binomial series expansion with $\alpha=-2$
\begin{align*}
(1+x)^\alpha=\sum_{n=0}^\infty\binom{\alpha}{n}x^n\qquad\qquad |x|<1\tag{1}
\end{align*}

We obtain
  \begin{align*}
\frac{1}{(7+x)^2}&=\frac{1}{7^2}\cdot\frac{1}{(1+\frac{x}{7})^2}\\
&=\frac{1}{7^2}\sum_{n=0}^\infty\binom{-2}{n}\left(\frac{x}{7}\right)^n\qquad\qquad |x|<7\tag{2}\\
&=\frac{1}{7^2}\sum_{n=0}^\infty\binom{n+1}{n}(-1)^n\left(\frac{x}{7}\right)^n\tag{3}\\
&=\sum_{n=0}^\infty\frac{(-1)^n(n+1)}{7^{n+2}}x^n\tag{4}\\
\end{align*}
  and the claim follows.

Comment:


*

*In (2) we apply the binomial expansion series with $\alpha=-2$ valid for $\left|\frac{x}{7}\right|<1$

*In (3) we use the binomial identity
$
\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q
$

*In (4) we use $\binom{n+1}{n}=\binom{n+1}{1}=n+1$
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