Sum of ((n^3)+9)/(n(3^n)
- Thread starter kilroymcb
- Start date
Don't leave your problem in the header.
\(\displaystyle \L\\\displaystyle\sum_{n=1}^{\infty}\frac{n^{3}+9}{n3^{n}}\)
I assume it is from 1 to infinity. You didn't say.
Try the ratio test:
\(\displaystyle \L\\\frac{(n+1)^{3}+9}{(n+1)3^{n+1}}\cdot\frac{n3^{n}}{(n+1)^{3}+9}=\frac{n}{3(n+1)}\)
That simplified nicely. Take the limit:
\(\displaystyle \L\\\lim_{n\to\infty}\frac{n}{3(n+1)}\)
If it's < 1, it converges.
\(\displaystyle \L\\\displaystyle\sum_{n=1}^{\infty}\frac{n^{3}+9}{n3^{n}}\)
I assume it is from 1 to infinity. You didn't say.
Try the ratio test:
\(\displaystyle \L\\\frac{(n+1)^{3}+9}{(n+1)3^{n+1}}\cdot\frac{n3^{n}}{(n+1)^{3}+9}=\frac{n}{3(n+1)}\)
That simplified nicely. Take the limit:
\(\displaystyle \L\\\lim_{n\to\infty}\frac{n}{3(n+1)}\)
If it's < 1, it converges.
