jcastel1
An archived instance of a Calc II forum
Homework 15 problem 3 part 1
mark
You’ll notice that the answer box looks like
\sum_{n=0}^{\infty} [???]
Your answer is the supposed to be the [???], which (in our notation) is the a_nx^n. Furthermore, they’ve already specified that they want a sum starting at n=0. So you you’ve got to specify your a_nx^n so that the first term corresponds to n=0, which you haven’t quite done.
I think it should go like so:
\frac{1}{1-x} \stackrel{x\to-x/3}{\longrightarrow} \frac{1}{1+\frac{x}{3}} \stackrel{\times 1/3}{\longrightarrow} \frac{1}{3+x} \stackrel{d/dx}{\longrightarrow} \frac{1}{(3+x)^2}.
Applying these same steps to the geometric series we get
\sum_{n=0}^{\infty} x^n \stackrel{x\to-x/3}{\longrightarrow} \sum_{n=0}^{\infty} (-1)^n \frac{x^n}{3^n} \stackrel{\times 1/3}{\longrightarrow} \sum_{n=0}^{\infty} (-1)^n \frac{x^n}{3^{n+1}} \stackrel{d/dx}{\longrightarrow} \sum_{n=1}^{\infty} (-1)^n n\frac{x^{n-1}}{3^{n+1}}.
This is almost what you’ve got, but I think you’ve got an error in your power of -1. Note, though, that I changed the starting index from n=0 to n=1 in the final step; we can do this since the zeroth term is zero (since we’re multiplying by n). When they say “shift the terms” they mean to add one to the n so that starting at zero in the new series is the same as starting at one in the old. Taking that into account, we get
\sum_{n=1}^{\infty} (-1)^n n\frac{x^{n-1}}{3^{n+1}} = \sum_{n=0}^{\infty} (-1)^{n+1} {n+1}\frac{x^{n}}{3^{n+2}}.
And that (without the summation symbol) should the answer.
jcastel1
This does not seem to work for the answer either.
mark
Did you type in my answer verbatim? Or did you include the parentheses that are obviously missing?? 
jcastel1
I did include the parentheses.