# Diverging and converging infinite series

1. Sep 16, 2010

### eptheta

I looked through some tutorials to find intervals of divergence and tests for divergence...
My series:
[PLAIN]http://img843.imageshack.us/img843/4193/51453212.jpg [Broken]
a and x are constants...
I did the ratio test and i get $$\rho$$=1, so i tried to apply the limit test to see if an is zero or does not exist... The problem is, i am not sure about my method...
Do i divide the numerator(1) by n also, if so, the test gives an=0..... But then again i am not sure...

Could someone please tell me if this sum is convergent or divergent (and the method you used to find this out)

Thanks

Last edited by a moderator: May 4, 2017
2. Sep 16, 2010

### Gerenuk

3. Sep 16, 2010

### mathman

Take x outside the summation. The terms in the sum behave like 1/n for large n. This series is well known divergent. The sum behaves like ln(n).

4. Sep 21, 2010

### eptheta

That makes sense, thanks.. I have one more question...

I was fiddling with my calculator and typed "Ans-(Ans-$$\sqrt{Ans}$$-2)" and kept clicking =...(The expression is in the same form as xn - f(x)/f'(x) but is just equal to $$\sqrt{Ans}$$+2....
I found that it converges as 4. It seemed logical enough....But then i decided to form the equation that represents f(x)...

so basically, what i got out of the expression is that y/(dy/dx)=x-$$\sqrt{x}$$-2
dy/y=dx/(x-$$\sqrt{x}$$-2)
Integrating(painfully), ln y=k(x-$$\sqrt{x}$$-2)($$\sqrt{x}$$-2/$$\sqrt{x}$$+2)1/3

Its root x -2 in the numerator not root 3 -2...

Where k is a constant... I should have tried with something simpler, but is this the correct way to form an approximation of a function that converges using Newton's method (or am i just being an oaf)
If it is the former, then how do i find k.
If it is the latter, then how do i find f(x) for :
[PLAIN]http://img714.imageshack.us/img714/5351/96407726.png [Broken]

Thanks.

Last edited by a moderator: May 4, 2017
5. Sep 24, 2010

### eptheta

Sorry, it is not summation to infinity, but rather term at infinity for the last expression.
-Thanks