# Need help for convergence/integrals..

• Dec 15th 2010, 07:12 AM
karagorge
Need help for convergence/integrals..
Hello i really need some help with these: (Headbang)

1. Find out if they are convergent - and if so find the SUM.
http://img839.imageshack.us/img839/9307/1stfb.png

2. Find the absolute/partial convergence or divergence:
http://img576.imageshack.us/img576/8536/2ndzr.png

3. Find the interval of convergence:
http://img716.imageshack.us/img716/3928/43932867.png

4. Find and draw the limit of the function:
http://img213.imageshack.us/img213/7530/4tht.png

5. Find out if the function has a limit and if so find the limit:
http://img219.imageshack.us/img219/5430/5tha.png

6. Find R, represent the double integral as iterated integral in two ways (vertical simple and horizontal simple):0
http://img18.imageshack.us/img18/60/7thk.png

7. Find the particular solution:
http://img704.imageshack.us/img704/61/8th.png

• Dec 15th 2010, 07:33 AM
FernandoRevilla
Only seven problems!

A little (very little help):

$\displaystyle\lim_{k \to{+}\infty}{\dfrac{1}{\sqrt[k]{e}}}=1\neq 0$

so,

$\displaystyle\sum_{k=1}^{+\infty}\dfrac{1}{\sqrt[k]{e}}$

is divergent.

Fernando Revilla
• Dec 15th 2010, 08:43 AM
FernandoRevilla
Another one:

$\displaystyle\lim_{k \to{+}\infty} \left(1+\dfrac{1}{2k+1}\right)^k=\ldots=\sqrt{e}\n eq 0$

so, the corresponding series is divergent.

Fernando Revilla

P.S. What have you tried for the rest?. This should not be a monologue.
• Dec 15th 2010, 08:58 AM
karagorge
sorry Fernando - My solutions for the first problem are the same
for the second problem - not much but i transformed it in http://img834.imageshack.us/img834/5906/unob.png
and stopped there. Now i'm on the third problem trying to solve it..
thanks for the help so far

------ just noticed k -> infinity ------ it shoud be k -> 1 -----
• Dec 15th 2010, 09:17 AM
karagorge
• Dec 15th 2010, 09:35 AM
FernandoRevilla
Quote:

Originally Posted by karagorge
sorry Fernando - My solutions for the first problem are the same
for the second problem - not much but i transformed it in http://img834.imageshack.us/img834/5906/unob.png

Take into account that $\sin k\pi=0$ so, the series

$\displaystyle\sum_{k=1}^{+\infty} \dfrac{\sin k\pi}{2}=\displaystyle\sum_{k=1}^{+\infty}0$

is convergent with sum $0$ .

Fernando Revilla
• Dec 15th 2010, 09:51 AM
FernandoRevilla
Quote:

Originally Posted by karagorge

That is right. Now, the series is absolutely convergent on $(-11/2,11/2)$ . You need to study the convergence at the end points of the interval.

(i) For $x=11/2$ you'll obtain:

$\dfrac{1}{22}\displaystyle\sum_{k=1}^{+\infty}\dfr ac{1}{n}$

divergent (harmonic series and algebra of series)

(ii) For $x=11/2$ you'll obtain:

$\dfrac{1}{22}\displaystyle\sum_{k=1}^{+\infty}\dfr ac{(-1)^n}{n}$

conditionally convergent (alternating harmonic series and algebra of series).

So, the interval of convergence is $[-11/2,11/2)$ .

Fernando Revilla
• Dec 15th 2010, 10:00 AM
FernandoRevilla
Quote:

Originally Posted by karagorge
4. Find and draw the limit of the function:
http://img213.imageshack.us/img213/7530/4tht.png

When $(x,y)\rightarrow \textrm{?}$

Fernando Revilla
• Dec 15th 2010, 10:08 AM
karagorge
Aaaa don't know - that might be a mistake - i'll ask tomorrow.
• Dec 15th 2010, 10:20 AM
FernandoRevilla
Quote:

Originally Posted by karagorge
Aaaa don't know - that might be a mistake - i'll ask tomorrow.

All right. Don't forget to show some work at every problem.

Fernando Revilla
• Dec 15th 2010, 10:32 AM
chiph588@
Quote:

Originally Posted by karagorge
5. Find out if the function has a limit and if so find the limit:
http://img219.imageshack.us/img219/5430/5tha.png

$x^4-y^4 = (x^2+y^2)(x^2-y^2)$ What does this tell you?

Quote:

7. Find the particular solution:
http://img704.imageshack.us/img704/61/8th.png
Assuming you meant $3y''+8y'-3y = 0$...

We must first solve $3\lambda^2+8\lambda-3 = 0$. This gives us $\displaystyle \lambda = \{-3,\tfrac13\}$.

Therefore $y = c_1e^{-3x}+c_2e^{x/3}$. Now use your conditions to solve for $c_1, c_2$.
• Dec 16th 2010, 05:20 AM
karagorge
Thanks for the fifth one - but i really can't figure out the 7th - Please finish it.. At the moment i'm solving other math problems so please forgive me for not answering on time or not including often in the process.
• Dec 16th 2010, 06:24 AM
karagorge