# Series and Sequences ex

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• Aug 16th 2010, 05:51 PM
fancynick
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.
• Aug 16th 2010, 06:03 PM
Vlasev
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.
• Aug 16th 2010, 06:30 PM
fancynick
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 ? )
• Aug 16th 2010, 06:44 PM
Vlasev
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.