Calculating the sum of a sequence

[goraemon]

Homework Statement

Compute $\sum\frac{4}{(-3)^n}-\frac{3}{3^n}$
as n begins from 0 and approaches infinity

Homework Equations

The Attempt at a Solution

I'm just getting started on sequences and series, and so far learned about the limit test, comparison test, arithmetic / geometric series and the like. Oh and a little bit about telescoping series. I'm just not sure how to approach this question. It doesn't look like a geometric or arithmetic series, and I tried to compute the first few terms to see whether it qualifies as a telescoping series, but it doesn't look like it. Thanks for any help.

 

[xiavatar]

Note that $\sum\frac{4}{(-3)^n}-\frac{3}{3^n}=\sum\frac{4}{(-3)^n}-\sum\frac{3}{3^n}=4\sum\frac{1}{(-3)^n}-3\sum\frac{1}{3^n}.$ Apply geometric sum formula.

 

[goraemon]

Note that $\sum\frac{4}{(-3)^n}-\frac{3}{3^n}=\sum\frac{4}{(-3)^n}-\sum\frac{3}{3^n}=4\sum\frac{1}{(-3)^n}-3\sum\frac{1}{3^n}.$ Apply geometric sum formula.

Ah ok...so then since n begins at zero...the answer would be...
$4\sum\frac{1}{(-3)^n}-3\sum\frac{1}{3^n}=4*\frac{1}{1-\frac{-1}{3}}-3*\frac{1}{1-\frac{1}{3}}=4*\frac{3}{4}-3*\frac{3}{2}=\frac{-3}{2}$

Is that right? Thanks!

 

[xiavatar]

Yes. That is correct.