How do you find the sum of sum from n=0 to infinity of (n(-2)^n)/(2^n). Do I use alternating series test and geometric series?
Follow Math Help Forum on Facebook and Google+
Originally Posted by Preston019 How do you find the sum of sum from n=0 to infinity of (n(-2)^n)/(2^n). Do I use alternating series test and geometric series? Note that $\displaystyle (-2)^n=(-1)^n2^n$ so $\displaystyle n\frac{(-2)^n}{2^n}=(-1)^nn$ So what?
Originally Posted by Plato Note that $\displaystyle (-2)^n=(-1)^n2^n$ so $\displaystyle n\frac{(-2)^n}{2^n}=(-1)^nn$ So what? It's divergent, right?
Yes, any series with terms that do not go to 0 is divergent.
View Tag Cloud