# series

• Nov 14th 2008, 12:24 PM
skabani
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.
• Nov 14th 2008, 12:55 PM
Mathstud28
Quote:

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

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

So the integral test applies, and note that $\int_2^{\infty}\frac{dx}{x\ln(x)}$ diverges.
• Nov 14th 2008, 01:03 PM
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?
• Nov 14th 2008, 01:30 PM
Mathstud28
Quote:

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 $\lim_{n\to\infty}\frac{\frac{1}{\sqrt{n}}}{\frac{1 }{\sqrt{n}\ln(n)}}=\infty$. Thus one of the series dominates the other, so we can glean nothing of its convergence from the LCT.
• Nov 14th 2008, 01:37 PM
skabani
ok

wait, so the original series converges then?
• Nov 14th 2008, 01:50 PM
Mathstud28
Quote:

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.