Convergence of series

**Darkprince** · March 20th 2011, 05:28 PM

I have to find if the following series converge:

a) Sum from k=1 to k=inf of k^k*e^(-k^2)

b) Sum from k=1 to k=inf of (1/sqrt(k))*e^(-2*sqrt(k))

For a I told that since k^k*e^(-k^2) is not equal to zero for k>=1 we can use the ratio test. I let a_k=k^k*e^(-k^2) then a_(k+1)/a_k=(((k+1)/k)^k)*(k+1) * e^(-2k+1) after some calculations. Then lim as k tends to inf of a_(k+1)/a_k is equal to 0<1 so our series converges absolutely by the ratio test.

Can anyone confirm my work and give me some guidance for which test should I use for part b? Thanks in advance!! Appreciate it!

c) I also have to determine the radius of convergence of the power series G(x)=Sum from k=1 to inf of (2k choose k)*x^k

For this I used the ratio test and found that abs(a_(k+1)/a_k)= 4 * abs(x) * (k+1/2)/(k+1) and as k tends to infinity then abs(a_(k+1)/a_k) tends to 4*abs(x)
Thus G(x) diverges for abs(x)>1/4 and converges absolutely for abs(x) < 1/4 so radius of convergence = 1/4

So basically if someone can check part a and c if I am correct and to guide me for part b.

Thanks in advance for any help!

**Prove It** · March 20th 2011, 06:37 PM

The ratio test is a good place to start.

a) $\displaystyle \sum_{k = 1}^{\infty}k^ke^{-k^2}$

$\displaystyle a_n = n^ne^{-n^2}$ and $\displaystyle a_{n+1} = (n+1)^{n+1}e^{-(n+1)^2}$

So $\displaystyle \lim_{n \to \infty}\left|\frac{a_{n+1}}{a_n}\right| = \lim_{n \to \infty}\left|\frac{(n+1)^{n+1}e^{-(n+1)^2}}{n^ne^{-n^2}}\right|$

$\displaystyle = 0$ according to http://www.wolframalpha.com/input/?i=limit+as+n+to+infinity+of+%28%28n%2B1%29^%28n%2 B1%29*Exp[-%28n%2B1%29^2]%29%2F%28n^n*Exp[-n^2]%29

Since this ratio $\displaystyle <1$ , the series is convergent.

**Darkprince** · March 20th 2011, 06:49 PM

Thank you! That's what I did with part a. Could you please help me with part b and confirm my work for part c? Thank you again!

**Darkprince** · March 21st 2011, 07:03 AM

I tried to do part b with the ratio test again and found that abs(a_(k+1)/a_k) = abs(sqrt(k/(k+1)) * e^ (2*(sqrt(k) * sqrt(k+1))
which tends to 1 as k tends to infinity, so I can't deduce if the sequence converges or diverges using the ratio test.

Thanks again for any help provided!

**Opalg** · March 21st 2011, 08:06 AM

R = 1/4 is correct for c).

For b), I would use the (limit) comparison test, remembering that "exponential growth is always faster than power growth".

The k'th term in b) is $\frac1{\sqrt k}e^{-2\sqrt k} = \frac1{\sqrt k\, e^{2\sqrt k}}.$ Now $e^{2\sqrt k}$ tends to infinity faster than any positive power of k. In particular, $e^{2\sqrt k} > k^{3/2}$ for all sufficiently large k. Therefore $\frac1{\sqrt k}e^{-2\sqrt k} < \frac1{k^2}$ (for all sufficiently large k) and so the series converges by comparison with $\sum\frac1{k^2}$ .

The comparison test is never popular with students, because it's not as "automatic" as the ratio test. You have to start by thinking about the order of magnitude of the k'th term of the series, and what to compare it with. But it's worth getting to grips with the comparison test, because it is actually the most useful of all the standard tests. For example, I would use it for part a) of this question, where it would get the result with less in the way of laborious calculation than the ratio test method. All you would need to say is that $k^ke^{-k^2} = e^{k\ln k}e^{-k^2} = e^{k(\ln k - k)}.$ But $\ln k-k<-1$ whenever k>1. Therefore $k^ke^{-k^2} < e^{-k}$ , and the series converges by comparison with the geometric series $\sum e^{-k}.$

**Darkprince** · March 21st 2011, 10:12 AM

Hello Opalg thanks for your reply. I have some questions to do. Assuming that exponential growth is faster than the power growth how does this leads us to e^(2*sqrt(k))> k^(3/2)?
Then you used that 1/(sqrt(k) * e^(2*sqrt(k)))< 1/(k^0.5 * k^(3/2))= 1/k^2 and then used the comparison test? Thanks for your reply, I am into your logic and solution the thing is that everything begins from e^(2*sqrt(k))>k^(3/2) which I didn't understand how you got it. Thanks for all the help, appreciate it!

I looked again your answer and you said that it tends to infinity faster than any positive power of k. The thing is that any exponent of e would do the job? i.e. if it was e^(3k^5) or e^(-sqrt(k)) would be tending to infinity faster than any positive power of k?

**Opalg** · March 21st 2011, 11:46 AM

Originally Posted by Darkprince

Assuming that exponential growth is faster than the power growth how does this leads us to e^(2*sqrt(k))> k^(3/2)?
Then you used that 1/(sqrt(k) * e^(2*sqrt(k)))< 1/(k^0.5 * k^(3/2))= 1/k^2 and then used the comparison test? Yes, that's right.
Thanks for your reply, I am into your logic and solution the thing is that everything begins from e^(2*sqrt(k))>k^(3/2) which I didn't understand how you got it.

The exact way in which "exponential growth is always faster than power growth" is as follows. Given any positive number n, it is always true that $e^x > x^n$ for all sufficiently large values of x. If you put $x = \sqrt k$ and n = 3/2 then you get $e^{\sqrt k} > k^{3/4}$ for all sufficiently large k. Now square both sides and you end up with $e^{2\sqrt k} > k^{3/2}$ for all sufficiently large k.

Originally Posted by Darkprince

I looked again your answer and you said that it tends to infinity faster than any positive power of k. The thing is that any exponent of e would do the job? i.e. if it was e^(3k^5) or e^(-sqrt(k)) would be tending to infinity faster than any positive power of k?

Yes, the same sort of argument as above will also show that e^(3k^5) (for example) tends to infinity faster than any positive power of k. It wouldn't work for e^(-sqrt(k)), because that is a negative exponential and it will tend to zero as k goes to infinity. (But it will go to zero faster than any negative power of k.)

**Darkprince** · March 21st 2011, 11:49 AM

Thank you very much Opalg, appreciate all your help!

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