Hi, I've run into some trouble trying to answer this power series question that asks:

Use multiplication or division of power series to find the first three nonzero terms in the Maclaurin series for the given function.

$\displaystyle e^{7x}ln(1-x/5)$

This is my work so far (all sums are n=0 to infinity):

$\displaystyle e^x$ = $\displaystyle \sum$ $\displaystyle \frac{x^n}{n!}$

$\displaystyle e^{7x}$ = $\displaystyle \sum$$\displaystyle \frac{7^nx^n}{n!}$

and

$\displaystyle

ln(1+x)$ = $\displaystyle \sum$ $\displaystyle (-1)^{n-1}\frac{x^n}{n}$

$\displaystyle ln(1-x/5)$ = $\displaystyle \sum$ $\displaystyle (-1)^{n-1}(-1)^n\frac{5^{-n}x^n}{n}$

$\displaystyle ln(1-x/5)$ = $\displaystyle \sum$ $\displaystyle (-1)^{2n-1}\frac{x^n}{n5^n}$

To find the first 3 non-zero terms, I think I'm supposed to do something like the following:

n=3 $\displaystyle \longrightarrow$ $\displaystyle a_0b_3+a_1b_2+a_2b_1+a_3b_0$

n=2 $\displaystyle \longrightarrow$ $\displaystyle a_0b_2+a_1b_1+a_2b_0$

n=1 $\displaystyle \longrightarrow$ $\displaystyle a_0b_1+a_1b_0$

n=0 $\displaystyle \longrightarrow$ $\displaystyle a_0+b_0$

What I got from doing that was:

($\displaystyle (-1)^{2n-1}$ becomes -1 for all n)

n=3

$\displaystyle -\frac{(7x)^0}{0!}*\frac{x^3}{3(5^3)}-\frac{(7x)^1}{1!}*\frac{x^2}{2(5^2)}-\frac{(7x)^2}{2!}*\frac{x^1}{1(5^1)}-\frac{(7x)^3}{3!}*\frac{x^0}{0(5^0)} $

= $\displaystyle -\frac{x^3}{375}-\frac{7x^3}{50}-\frac{49x^3}{10}-\frac{343x^3}{0}$

= $\displaystyle -\frac{3781x^3}{375}-\frac{343x^3}{0}$

n=2

$\displaystyle -\frac{(7x)^0}{0!}*\frac{x^2}{2(5^2)}-\frac{(7x)^1}{1!}*\frac{x^1}{1(5^1)}-\frac{(7x)^2}{2!}*\frac{x^0}{0(5^0)}$

= $\displaystyle -\frac{x^2}{50}-\frac{7x^2}{5}-\frac{49x^2}{0}$

= $\displaystyle -\frac{71x^2}{50}-\frac{49x^2}{0}$

n=1

$\displaystyle -\frac{(7x)^0}{0!}*\frac{x^1}{1(5^1)}-\frac{(7x)^1}{1!}*\frac{x^0}{0(5^0)}$

= $\displaystyle -\frac{x}{5}-\frac{7x}{0}$

n=0

$\displaystyle -\frac{(7x)^0}{0!}-\frac{x^0}{0(5^0)}$

$\displaystyle -1-\frac{1}{0}$

How do I get my 3 terms from the above? I have a lot of fractions where I'm dividing by 0.

Edit: I just realized that I changed $\displaystyle e^{7x}$ into a power series, would the Maclaurin series for that function be the same?