- #1
ognik
- 643
- 2
Use the comparison test to see if \(\displaystyle \sum_{1}^{\infty}{\left[n\left(n+1\right)\right]}^{-\frac{1}{2}} \)converges?
I tried \(\displaystyle n+1 \gt n, \therefore n(n+1) \gt n^2 , \therefore {\left[n(n+1)\right]}^{\frac{1}{2}} \gt n, \therefore {\left[n(n+1)\right]}^{-\frac{1}{2}} \lt \frac{1}{n}\) - no conclusion possible here
I tried \(\displaystyle n+1 \gt n, \therefore n(n+1) \gt n^2 , \therefore {\left[n(n+1)\right]}^{\frac{1}{2}} \gt n, \therefore {\left[n(n+1)\right]}^{-\frac{1}{2}} \lt \frac{1}{n}\) - no conclusion possible here