1. Feb 2, 2008

### frixis

1. The problem statement, all variables and given/known data
determine whether the sequence converges or diverges and find its limits
1$(1+\frac{1}{n})^n$
2 cos$$\pi$$n
the overbar indicates that the digits underneat repeat indefinitely. Express as a series and find the rational number(since i cant find the latex for overbar, the brackets represent the overbar)
0.(23)
3.2(394)
show divergence or convergence by the nth term text
$\displaystyle\sum_{n=2}^{\infty} \frac{1}{n\sqrt{{n^2}-1}}$
use a basic comparision test to determine convergence
$\displaystyle\sum_{n=1}^{\infty} \frac{\arctan n}{n}$
determine convergence
1$\displaystyle\sum_{n=1}^{\infty} \frac{2n+n^2}{n^3+1}$
2$\displaystyle\sum_{n=1}^{\infty} \frac{n^2+2^n}{n+3^n}$
3$\displaystyle\sum_{n=1}^{\infty} \frac{\mathrm{ln} n}{n^3}$
4 $$\displaystyle\sum_{n=2}^{\infty} {\frac{1}{{n^3}{\sqrt[3]{\mathrm{ln}n}}}$$
5$\displaystyle\sum_{n=1}^{\infty} n{\tan{\frac{1}{n}}$
6$\displaystyle\sum_{n=1}^{\infty} {(1+\frac{1}{n})}^n$
7 $1+\frac{1\cdot 3}{2!}+\frac{1\cdot 3\cdot 5}{3!}+....+\frac{1.3.5....(2n-1)}{n!}+....$
find every real no k for which the series converges
$\displaystyle\sum_{n=2}^{\infty} \frac{1}{n^k{\mathrm{ln}n}}$

use the proof of the integral test to show that, for every positive int n >1
${\mathrm{ln}(n+1)}{<1+\frac{1}{2}+\frac{1}{3}+......+\frac{1}{n}<1+{\mathrm{ln}n}$

determine wheter the series is absolotely convergent, conditionally convergent or divergent.
$\displaystyle\sum_{n=1}^{\infty} {(-1)}^n \frac{\cospin}{n}$

show that the alternating series converges for ever positive integer k
$\displaystyle\sum_{n=1}^{\infty} {(-1)}^n \frac{{(\mathrm{ln}n)}^k}{n}$

find the interval of convergence of the power series
1 $\displaystyle\sum_{n=1}^{\infty} {(-1)}^n \frac{x^n}{\sqrt{n}}$ can do the whole getting to the interval just cant decide whether or not it converges or diverges at -1 and 1 which are the intervals endpoints
2 $\displaystyle\sum_{n=0}^{\infty} {\frac{n!}{100^n} x^n}$ issues with endpoints again
3 $\displaystyle\sum_{n=0}^{\infty} \frac{x^{2n+1}}{(-4)^n}$

find the radius of convergence of the power series for positive integers c and d
$\displaystyle\sum_{n=0}^{\infty} {\frac{{(n+c)}!}{n!{(n+d)}!} x^n$

if the interval of convergence of $\displaystyle\sum {a_n}{x^n}$ is (-r,r], prove that the series is conditionally convergent at r

3. The attempt at a solution
um.. okay so i have to put down the attempts to all the questions? its a culmination of all issues...
for the decimals ones... i did the whole sum = a/1-r thing... for some reason i was doing something wrong...
for all convergence ones i always tried the ratio test or the root test... but i still had issues with the convergence... specially when ln comes in... i dont know y...
and the last one which is the most recent... i deduced that the ratio of the n+1 term and the n term is 1/r but then i tried putting that in the equation and i was pretty lost... cuz u'd get 1 and... yeah... okay...
so if i need to post all my attempts please let me know...

Last edited: Feb 2, 2008
2. Feb 2, 2008

3. Feb 2, 2008

### Gib Z

You have to post attempts to every single question. I suggest you give them a harder try, reduce the list of questions you need help with because you seem to have given us the entire list of questions, not the specific ones you need help with.

For every question you show us a decent attempt at, we will give you help on. So try to reduce the list first.

4. Feb 2, 2008

### mjsd

perhaps just show us your attempt on one or two of these and then we may be able to point out something that is confusing you or gaps in your knowledge.

5. Feb 2, 2008

### nuclearrape66

i'll work them out and if you still need help just message me...=)

6. Feb 2, 2008

### Gib Z

No, that is not in accordance with forum rules. If he wants help, he is going to post his work on this thread, not message you to get some solutions.

7. Feb 2, 2008

### nuclearrape66

private messaging has nothign to do with forum posting...understand these rules....

8. Feb 2, 2008

### Gib Z

So you are going to hand out answers to someone who obviously needs help on how to do them? Wow, that helps him learn.

Sorry, but I am seriously not going to watch you stuff up this guys understanding even further, hes already behind in his class and your going to make it worse.