1. ## series

Converge or Diverge?

I'm not sure how to go about solving these...help?

1. Sigma (Sqrt(2n+1) - Sqrt(2n))
n = 1, and goes to infinity

1. Sigma (1/(Sqrt(n)ln (n)))
n= 2, and goes to infinity

thanks.

2. Originally Posted by skabani
Converge or Diverge?

I'm not sure how to go about solving these...help?

1. Sigma (Sqrt(2n+1) - Sqrt(2n))
n = 1, and goes to infinity

1. Sigma (1/(Sqrt(n)ln (n)))
n= 2, and goes to infinity

thanks.
1.Consider the sequence of partial sums for the first one. Do you notice a pattern?

2.$\displaystyle \sum_{n=2}^{\infty}\frac{1}{\sqrt{n}\ln(n)}>\sum_{ n=2}^{\infty}\frac{1}{n\ln(n)}$

Now consider that if $\displaystyle a_n=\frac{1}{n\ln(n)}$, then $\displaystyle \forall{n}\in[2,\infty)~a_n>0\wedge{a_n\in\mathcal{C}}\wedge{a_n \in\downarrow}$

So the integral test applies, and note that $\displaystyle \int_2^{\infty}\frac{dx}{x\ln(x)}$ diverges.

3. thanks.

could we also use the limit comparison test and compare the series to 1/Sqrt(n), where c>0, and the series would diverge?

4. Originally Posted by skabani
thanks.

could we also use the limit comparison test and compare the series to 1/Sqrt(n), where c>0, and the series would diverge?
No, because $\displaystyle \lim_{n\to\infty}\frac{\frac{1}{\sqrt{n}}}{\frac{1 }{\sqrt{n}\ln(n)}}=\infty$. Thus one of the series dominates the other, so we can glean nothing of its convergence from the LCT.

5. ok

wait, so the original series converges then?

6. Originally Posted by skabani
ok

wait, so the original series converges then?
As I showed with the integral test the comparison series divergres, which implies that since the original series is greater than the comparison series it diverges.