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.

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- Nov 14th 2008, 12:24 PMskabaniseries
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. - Nov 14th 2008, 12:55 PMMathstud28
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. - Nov 14th 2008, 01:03 PMskabani
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? - Nov 14th 2008, 01:30 PMMathstud28
- Nov 14th 2008, 01:37 PMskabani
ok

wait, so the original series converges then? - Nov 14th 2008, 01:50 PMMathstud28