series help!
- Thread starter clw89
- Start date
Try using the integral test.
\(\displaystyle \int_{1}^{L}\frac{ln(n)}{n^{3}}=\frac{-ln(L)}{2L^{2}}-\frac{1}{4L^{2}}+\frac{1}{4}\)
Now, take the limit:
\(\displaystyle \lim_{L\to\infty}\left[\frac{-ln(L)}{2L^{2}}-\frac{1}{4L^{2}}+\frac{1}{4}\right]\)
See the limit?. What does that tell you about convergence?.
\(\displaystyle \int_{1}^{L}\frac{ln(n)}{n^{3}}=\frac{-ln(L)}{2L^{2}}-\frac{1}{4L^{2}}+\frac{1}{4}\)
Now, take the limit:
\(\displaystyle \lim_{L\to\infty}\left[\frac{-ln(L)}{2L^{2}}-\frac{1}{4L^{2}}+\frac{1}{4}\right]\)
See the limit?. What does that tell you about convergence?.
