
Series and Sequences ex
Hello,
I'm preparing for my mathematics exams and am a little bit struggling with series and functions. Since my closest lecturer is 3000km away, decided to ask some help online:)
The following Series are convergent or divergent?
http://img69.imageshack.us/img69/922...otestexam2.jpg
http://img231.imageshack.us/img231/2...iotestexam.jpg
I solved it but really not sure if it is right.
And was struggling to solve this one for Sequences: whether or not this sequence converge, if does state limit: 2010^n / n^2010.

The second one is correct. Now for the first one
if $\displaystyle a_n = \frac{(1)n}{\sqrt{n}} = \sqrt{n}$ , then it does not converge since the term oscillates more an more wildly.
If $\displaystyle a_n = \frac{(1)^n}{\sqrt{n}}$, then you can just use the alternating series test. This term goes to 0 as n goes to infinity, so we have that the series converges! The work you have shown is incorrect though, because you cannot factor 1 out of the absolute value. The limit is indeed 1, so it gives no information on convergence.

Thank You for the explanation and sorry for confusing you, it was (1)^n in the example  bad paint skills!
Maybe have any ideas on the bottom exercise, the sequence 2010^n / n^2010 diverge ir converge ( limit ? )

The numerator is an exponential while the denominator is a polynomial. At first this ratio is very small for small numbers. However, when n = 2010, the ratio is 1. After that the numerator wins.
If you compare consecutive terms you have:
$\displaystyle \displaystyle \frac{a_{n+1}}{a_n} = \frac{2010^{n+1}}{(n+1)^{2010}}\frac{n^{2010}}{201 0^n} = 2010\left(\frac{n}{n+1}\right)^{2010}$
The limit of this is of course 2010 and since it's much larger than 1, the sequence diverges to infinity.