# Thread: Calc 2: Converge/Diverge

1. ## Calc 2: Converge/Diverge

How do i find out if these converge/diverge:

inifinity
SIGMA sqrt(2n+1)/(n^2)
n=1

infinity
SIGMA (3^K+K)/(K!)
k=1

I dont know what K! is and i dont know what to do with the square root. It stumps me and I dont know what theorems to use. The integral test? Bounded sum test? p-series test?

2. Originally Posted by MathNeedy18
How do i find out if these converge/diverge:

inifinity
SIGMA sqrt(2n+1)/(n^2)
n=1

infinity
SIGMA (3^K+K)/(K!)
k=1

I dont know what K! is and i dont know what to do with the square root. It stumps me and I dont know what theorems to use. The integral test? Bounded sum test? p-series test?
$\displaystyle 0 < \frac{\sqrt{2n+1}}{n^2} < \frac{\sqrt{2n+2n}}{n^2} = \frac{2\sqrt{n}}{n^2} = \frac{2}{n^{3/2}}$ .......

For the second one, have you tried the ratio test?

3. Thanks Mr. Fantastic. My math teacher has been zooming through chapters cuz we are weeks behind and therefore just writes down definitions and what not and doesnt do examples. I learn from examples so I'm not even quite sure how to do the ratio test

4. Originally Posted by MathNeedy18
Thanks Mr. Fantastic. My math teacher has been zooming through chapters cuz we are weeks behind and therefore just writes down definitions and what not and doesnt do examples. I learn from examples so I'm not even quite sure how to do the ratio test
Does your book have a chapter or section that explains the ratio test and gives examples .....?

5. Hi,

Often remember to thank Riemann ^^

We call the ratio test "d'Alembert test" in France, if it can help...

If the general term is An, if $\displaystyle \frac{A_{n+1}}{A_n} < 1$, the serie converges. If > 1, it diverges. If =1, you have to try

6. Originally Posted by Moo
Hi,

Often remember to thank Riemann ^^

We call the ratio test "d'Alembert test" in France, if it can help...

If the general term is An, if $\displaystyle \frac{A_{n+1}}{A_n} < 1$, the serie converges. If > 1, it diverges. If =1, you have to try
The ratio test will be one of your best friends in proving the convergence or divergence of a series. Other tests that I found most useful were the comparison test, root (although it fails when ratio test fails) and limit comparison test.

7. Originally Posted by colby2152
root (although it fails when ratio test fails) and
This is WRONG. The root test is SUPERIOR to the ratio test.

The actual full statement is based on limsup and liminf. The test says given $\displaystyle \sum_{n=0}^{\infty}a_n x^n$. Consider $\displaystyle L = \limsup |a_n|^{1/n}$, the beauty here is that this limit exists (or $\displaystyle L=\infty$) not matter what, the reciprocal will be the radius of convergence. The ratio test is not superior, for one thing we require that the terms $\displaystyle a_n$ be non-zero to use this test! While in root test this is not a necessity! Furthermore, the ratio test is based on root test, because of the fact that: $\displaystyle \liminf |a_{n+1}/a_n| \leq L \leq \limsup |a_{n+1}/a_n|$. Thus, if $\displaystyle \limsup |a_{n+1}/a_n| < 1$ then $\displaystyle L<1$ and we have convergence. If $\displaystyle \liminf |a_{n+1}/a_n| > 1$ then $\displaystyle L>1$ and we have divergence. All because of the root test. And finally we can give examples where there ratio test fails and root test succedes.

8. Originally Posted by ThePerfectHacker
This is WRONG. The root test is SUPERIOR to the ratio test.

The actual full statement is based on limsup and liminf. The test says given $\displaystyle \sum_{n=0}^{\infty}a_n x^n$. Consider $\displaystyle L = \limsup |a_n|^{1/n}$, the beauty here is that this limit exists (or $\displaystyle L=\infty$) not matter what, the reciprocal will be the radius of convergence. The ratio test is not superior, for one thing we require that the terms $\displaystyle a_n$ be non-zero to use this test! While in root test this is not a necessity! Furthermore, the ratio test is based on root test, because of the fact that: $\displaystyle \liminf |a_{n+1}/a_n| \leq L \leq \limsup |a_{n+1}/a_n|$. Thus, if $\displaystyle \limsup |a_{n+1}/a_n| < 1$ then $\displaystyle L<1$ and we have convergence. If $\displaystyle \liminf |a_{n+1}/a_n| > 1$ then $\displaystyle L>1$ and we have divergence. All because of the root test. And finally we can give examples where there ratio test fails and root test succedes.
Whoops, I got it backwards... the ratio test fails when the root test fails...

9. Never heard 'bout the superiority of one test upon another... The two have always been equivalent to us