Radius of Convergence, Sum [1...inf] x^sqrt(x)??

[Zan4]

Help 4 woman struggling with starting steps? Radius of Conv, Sum [1...inf] x^sqrt(x)?

I am pretty sure I should start with investigating:

lim (x=inf) x^sxrt(x) / (x+1)^sqrt(x+1)

But the problem is I dont know how to get sqrt´s down or eliminated to get further.

Can you help me?

 

[stapel]

I am pretty sure I should start with investigating:

What are you "starting"? Or are you saying that your question is in the subject line?

 

[Zan4]

What are you "starting"? Or are you saying that your question is in the subject line?

Thank you for making 1st comment!


Radius of Convergence of Sum [1...inf] x^sqrt(x) is what I am aiming to solve.

And I have learnt that I should start solving by investigating

lim (x=inf) |A_(n) / A_(n+1)|


So I would like to get some hints that would help me to solve this lim.

And STRICT question is, how to go on with this:

lim (x=inf) (x^sqrt(x)) / ((x+1)^(sqrt(x+1))


I apologize if I was unclear.

 

[HallsofIvy]

Thank you for making 1st comment!


Radius of Convergence of Sum [1...inf] x^sqrt(x) is what I am aiming to solve.

And I have learnt that I should start solving by investigating

lim (x=inf) |A_(n) / A_(n+1)|

Yes, that's the "ratio" test.

So I would like to get some hints that would help me to solve this lim.

And STRICT question is, how to go on with this:

lim (x=inf) (x^sqrt(x)) / ((x+1)^(sqrt(x+1))


I apologize if I was unclear.

Since numerator and denominator go to infinity separately, you might try "L'Hopital's rule", differentiating numerator and denominator separately.

Another thing you could try is the "root test", a logarithmic variation on the ratio test:
"If \(\displaystyle \lim_{x\to\infty} \sqrt[x]{a_x}< 1\) then the series \(\displaystyle \sum a_x\) converges"

 

[Zan4]

Yes, that's the "ratio" test.



Since numerator and denominator go to infinity separately, you might try "L'Hopital's rule", differentiating numerator and denominator separately.

Another thing you could try is the "root test", a logarithmic variation on the ratio test:
"If \(\displaystyle \lim_{x\to\infty} \sqrt[x]{a_x}< 1\) then the series \(\displaystyle \sum a_x\) converges"

I still thank you for your answers! You noted very rational things. I finally solved this by noticing I can make use of e. My solving here, if you are eager to read it through:

My sentence in beginning: [(x^sqrt(x)) / ((x+1)^sqrt(x+1))].
I changed it to be: [(e^(sqrt(x)*lnx)) / (e^(sqrt(x+1)*ln(x+1)))]
which equals to e^[(sqrt(x)*lnx)-(sqrt(x+1)*ln(x+1))]

Now I can use marking where b_x= [(sqrt(x)*lnx)-(sqrt(x+1)*ln(x+1))] (it is exponent from previous line)

I make additive expantion to make b_x look like this:
b_x=[(sqrt(x)*lnx)-(sqrt(x)*ln(x+1))+(sqrt(x)*ln(x+1))-(sqrt(x+1)*ln(x+1))]

I get that previous sentence equals to
[sqrt(x)*(-ln(1+1/x))]+[(sqrt(x)-sqrt(x+1))*ln(x+1)] which converges to 0

Finally e^b_x converges to 1
and Radius of Convergence is R=1.