power series representation of a function by differentiation

Question

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.

Answer 1

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 $\binom{n+1}{n}=\binom{n+1}{1}=n+1$