Can a finite series like this be evaluated?

Hertz

I came across this series by recognizing a pattern while trying to evaluate an integral. I was wondering if the series could be solved in a generalized form where n can vary, and if so, can the limit then be taken as n approaches infinity?

You can't take the infinite series without first solving the finite series because of the (n - k)!
:S!

[itex]\Sigma^{n}_{k = 0}((\frac{n!}{(n - k)!})x ln^{n - k}(x))[/itex]

chiro

Hey Hertz.

One thing you should do if you want at least see if you can calculate the limit as n goes to infinity is to check for convergence. If the series doesn't converge then you have your answer for the infinite case.

Are you aware of the convergence tests used for power series?

Hertz

Yes, the infinite series diverges for all x and for all n. I believe it's because of the (n - k)!. Obviously, (n - k)! can't be evaluated unless k is less than n, so we can't let k approach infinity; k must have an upper bound of n.

The only way I can see to evaluate this series is to solve the finite series with n as the upper limit and THEN take the limit as n approaches infinity.

chiro

Ok no worries.

As for the simplifying it, why not try putting the equation into the (a + b)^n form since you have a binomial coeffecient and you can take the x outside of the sigma since the dummy variable is not used: (hint: try x(1 + a)^n for a relevant a term).

Hertz

Hmm.. All I can seem to do is get it to this:

[itex]x\Sigma^{n}_{k = 0}((\frac{n!}{(n - k)!})(\frac{ln^n(x)}{ln^k(x)}))[/itex]

chiro

Let a = ln(x). Is there a relationship between your summation and x(1 + a)^n? (Have you ever come across the binomial theorem? http://en.wikipedia.org/wiki/Binomial_theorem)

Hertz

Ah! The binomial theorem is actually the next section I need to read about in my calc 2 book. I'll go ahead and do that before I revisit this thread :) Thanks for the help

Hertz

Ok, so here's what I have so far:

[itex]x\Sigma^{n}_{k = 0}(k!(^{n}_{k})(\frac{a^n}{a^k}))[/itex]

[itex]\frac{n!}{(n - k)!} = k!(^{n}_{k})[/itex]

As you may have guessed, I don't have much experience with binomial coefficients, but I'll continue to play around with this anyways and see if I can work it into something I know how to evaluate. You were hinting that I should try to reform it into (1 + x)^n? I'll look into that but I don't see how I would do that quite yet with those different powers of a and the k! in there.

[Edit]
One other question:

I have the series in the form:
[itex]x\Sigma^{n}_{k = 0}((^{n}_{k})(\frac{(k!a^n)^{1/k}}{a})^k)[/itex]

Can I convert it to this even though the second factor has n's and k's in it?
[itex]x((\frac{(k!a^n)^{1/k}}{a})^k) + 1)^n[/itex]

chiro

Take a look at the wiki page I posted earlier: Look at this part - http://en.wikipedia.org/wiki/Binomia...of_the_theorem and look at the second expansion. Let the x part on this page = 1 and the y part = a and compare the summation you have with the summation on the wiki page.

haruspex

I get the asymptotic form x²n! - x lnⁿ⁺¹(x)/(n+1) + smaller terms. There cannot be a closed form since it involves the truncated expansion of exp(a).

Hertz

Doing what you said I have this:

[itex]\Sigma^{n}_{k = 0}k!(^{n}_{k})y^{n - k}[/itex]

The only problem is I have a k! instead of an [itex]x^k[/itex].

chiro

Sorry! Yeah you can't put it in the binomial (I forgot about the extra k!) so yeah sorry :(

You'll have to do what haruspex said.

Hertz

Shoot! :S! Hmm, he didn't really say much though. Just partially said what he got. I would be happy to know where it came from though :)

A bit of good news at least, after changing the series to binomial form it will now converge for all x and n. However, it's not useful unless I can actually take the limit as n goes to infinity so hopefully another opportunity to get that sigma out of there will come soon :S

haruspex

True.
- take the x and n! outside the sum
- switch k to n-k everywhere
- if a = ln(x), what we have left in the sum is the truncated expansion of e^a. So it evaluates to e^a - Ʃ(a^k/k!), summing for k > n. I took the leading term of that sum and threw away the rest as being smaller order.
e^a = x, so we end up with x²n! as the dominant term.

romsofia

It might be easier if you posted the integral you're having trouble evaluating, as this could be the wrong approach.

Hertz

How come I can switch k to n - k? Doesn't that change the meaning of the expression?

Ideally I'd like to be able to integrate [itex]x^nln^n(x)[/itex] but for now I'm just trying to integrate [itex]ln^n(x)[/itex]

e-
With respect to x obviously