# Thread: test for convergence of series

1. ## test for convergence of series

sigma from 1 to infinite of 1/[(squareroot n+1) -nsquareootn]

2. Originally Posted by twilightstr
sigma from 1 to infinite of 1/[(squareroot n+1) -nsquareootn]
Do you mean $\displaystyle \sum_{n=1}^\infty \frac{1}{\sqrt{n+1}- n\sqrt{n}}$?

An obvious first step would be to multiply numerator and denominator by $\displaystyle \sqrt{n+1}+ n\sqrt{n}$ so that each term is $\displaystyle \frac{\sqrt{n+1}+ n\sqrt{n}}{n+1- n^3}$. Now a comparison test should do it.

Or, now that I think about it, just a comparison to $\displaystyle \frac{1}{n^{3/2}}$!

3. no. 1/[(squareroot (n+1)) - (squareroot (n-1))]

4. Originally Posted by twilightstr
no. 1/[(squareroot (n+1)) - (squareroot (n-1))]
If you mean

$\displaystyle \sum_{n=1}^{\infty} \frac{1}{\sqrt{n+1}-\sqrt{n-1}}$

then HallsofIvy's suggestion still works - multiply btop and bottom by $\displaystyle \sqrt{n+1}+\sqrt{n-1}$

giving

$\displaystyle \sum_{n=1}^{\infty} \frac{\sqrt{n+1}+\sqrt{n-1}}{2}$

which diverges (terms don't go to zero)

You could also do a limit comparison with $\displaystyle \sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}$

5. $\displaystyle \frac{1}{\sqrt{n+1}-\sqrt{n-1}} \ge \frac{1}{\sqrt{n+1}}$

Since the second is divergent (You should be able to figure that $\displaystyle \sum_{n=1}^{\infty} \frac{1}{\sqrt{n+1}}$ is divergent). By the comparison test, $\displaystyle \sum_{n=1}^{\infty}\frac{1}{\sqrt{n+1}-\sqrt{n-1}}$ must also diverge.