Testing for convergence/divergence

[Dr Zoidburg]

I've got a couple of problems I'm stuck on. Any help gratefully received!

Test for convergence/divergence:
\sum_{n=1}^{\infty} \frac {(n+1)}{n^3 ln(n+2)}

What test should I do here? Can I rearrange the equation to be:
\frac{(\frac{1}{n^2} + \frac{1}{n^3})}{ln(n+2)}

and then use L'Hopital's Rule?
Or do I need to use the Integral test? If so, what should I put u=?

Next Question:
For what values of k isthe following series absolutely convergent? For what values k>=0 conditionally convergent?
\sum_{n = 3}^{\infty}\frac{(-1)^n}{n(\ln{n})[\ln{(\ln{n})}]^k}

I've spent a looooonnnng time on this! This is what I've got thus far:
make u= ln(ln(n))
du=\frac{1}{n(ln(n))} dn
And then integrate as follows:
\lim_{b = 3 \to \infty} \int_{3}^{b} \frac{1}{u^k} du
which gives us:
\frac{u^{(1-k)}}{1-k}|_{3}^{b}
So does this mean it's convergent for k>1 and divergent for k=1?

Last question:
prove:
\lim_{n \to \infty} \frac {n^p}{a^n} = 0 where a & p are constants and a>1 & p>0
I've tried using the Ratio test here, but I ended up with 1/a, which obviously is not 0.

Thanks for any and all help/advice!

 

[Brilliant]

Hey, I'm just learning this stuff too but I'll tell you what I think.

So to determine when this converges or diverges.
<br /> \sum_{n=1}^{\infty} \frac {(n+1)}{n^3 ln(n+2)}<br />
I actually think of ratio test on this one.
So
<br /> \lim_{n \to \infty} \frac {(n+2)}{(n+1)^3 ln(n+3)} * \frac {n^3 ln(n+2)}{(n+1)}<br />
will simplify to
<br /> \lim_{n \to \infty} \frac {(n^4+2n^3)ln(n+2)}{(n^4...)ln(n+3)}<br />
It seemed to me that the ln()'s weren't going to affect the limit and the greatest powers on top and bottom are both 4 with coefficients of 1, so the limit to infinity would just be 1. The ratio test says to converge the absolute value of this limit must be less than 1. I conclude that it does not converge. The only thing that could be wrong is how I ignored the ln(). But I mean they are both essentially ln(n) on top and bottom.

On your last question
<br /> \lim_{n \to \infty} \frac {n^p}{a^n} = 0<br />
I would just think of it with L'Hopital's Rule

As you take more and more derivatives n^p will go to zero, because p is constant and it will be reduced with each dirv. And a^n will just be ln(a)a^n. Or maybe that's log(a) Anyway, the bottom will be large and the top will be 0.


You're second question doesn't look fun. I would consider the alt. series test. So, An must be less than A(n+1). That seems like it will be true since incrementing n, increases the denominator. The other part of the alt. series test says that the limit to inf. of An must be 0. Well I think that will always happen if k is positive, but if k were negative then you could move the that ln(ln(n)) to the top, then who knows if it would still go to 0.

I hope something I wrote can help. I can be pretty terrible at explaining things. Also I don't claim that any of my suggestions will work

Let me know if you need clarification.

 

[jbunniii]

Another, perhaps easier, way to do the first part:

\frac{n+1}{n^3 \ln (n+2)} = \frac{n}{n^3 \ln (n+2)} + \frac{1}{n^3 \ln (n+2)} \leq \frac{n}{n^3} + \frac{1}{n^3} = \frac{1}{n^2} + \frac{1}{n^3}

and use the comparison test.

 

[Brilliant]

That makes much more sense. Also the thing I did with the radio test failed. I forgot that if it equals 1, then it is inconclusive.

Nice job jbunni.

 

[Dr Zoidburg]

Thanks for the replies. It's helped me get my head around these problems better.
I like jbunniii's solution. very clever and succinct.
For this question:
\lim_{n \to \infty} \frac {n^p}{a^n} = 0
my solution is:
Apply L'Hopital's Rule:
\frac{p}{ln(a)} [\lim_{n \to \infty} \frac {n^{p-1}}{a^n}]
We continue applying L'Hopital's Rule until we eventually get to:
\frac{p!}{ln^p(a)} [\lim_{n \to \infty} \frac {1}{a^n}]
Which gives us 0, as
\lim_{n \to \infty} \frac {1}{a^n} = \frac{1}{\infty} = 0

That's another assignment out of the way.
Happy Easter all!