Is there a formula for

Also, I have the conjecture that

Hmm. It does seem to be divisible by 3 in all the examples my computer has produced so far. (Note that if x=14 the sum is 3 x 3547114323481, and the latter is prime.)

There does exist a neat proof that it is always an integer.

We have $Q(f)=\sum _{n=1}^{\infty }n/z^n$, $Q^2(f)=\sum _{n=1}^{\infty }{n}^2/z^n$, etc.

(An operator is a sort of meta-function which acts on functions. So differentiation is an operator, as is taking Fourier transforms and all sorts of other weird notions.)

So your series is $1/2Q^{x}f(z)$ evaluated at $z=1/2$.

Now consider the terms in the sequence of functions $Q^k(f)$. The first is $z/(1-z{)}^2$, the second $z(1+z)/(1-z{)}^3$, etc.

All take the form $P(z)/(1-z{)}^{k+1}$, where $P(z)$ is a polynomial in $z$ with integer coefficients and degree at most $k$. We shall prove this by induction on $k$. It is true for $k=1$ and 2 (work these out if you want).

Suppose $Q^k(f)$ is of this form. Then $Q^{k+1}(f)=$

$zd/\mathrm{dz}{Q}^k(f)$

$=zd/\mathrm{dz}(P(z)/(1-z{)}^{k+1})$

$=z(P\text{'}(z)(1-z{)}^{k+1}+(k+1)P(z)(1-z{)}^k)/(1-z{)}^{2k+2})$

$=z((1-z)P\text{'}(z)+(k+1)P(z))/(1-z{)}^{k+2}$

Since $P(z)$ is of degree $k$, as is $(1-z)P\text{'}(z)$, and both have integer coefficients, the numerator of this has integer coeffs and is of degree at most $k+1$, as required. (We can in fact prove, by an easy extension of this induction, that it is of degree precisely $k$ and has leading coefficient 1; but this is not relevant.)

Anyway, let's now put $z=1/2$. We have $\sum _{n=1}^{\infty }{n}^k/2^{n+1}=1/2Q^k(f)$ with $z=1/2$. This is $(1/2P(1/2))/((1/2{)}^{k+1})$.

Now, the polynomial $P(1/2)$ is a sum of terms all of the form $a(i)/2^i$ where $i$ is at most $k$, as the polynomial itself is of degree at most $k$. So when the numerator is multiplied by $2^{k+1}$, this is $2^{k}P(1/2)$ and all of the terms become integers, since the $a(i)$ are integers. So $2^{k+1}$ times the numerator is an integer.

However, the denominator is $1/2^{k+1}$, so this value - $2^{k+1}$ times the numerator - is the value of the fraction. Which is therefore an integer. QED.

(You may like to extend this to show that it is an odd integer; you will need the fact that the leading coefficient of $P(z)$ is precisely 1.)
I will have a good look at the "divisible by 3" bit!

David

I just noticed if you do a bit of manipulation, then Brad's sum (the one in the second message) is also given by the finite double summation:

And it also follows pretty quickly from the double sum that it is an odd integer. (The only odd terms in the double sum occur when n = x-m.)

Now for multiples of 3. If x is even then n^x = 0 (mod 3) if n is a multiple of 3, and n^x = 1 (mod 3) otherwise. Reducing the sum mod 3 we get:

There's a sign error in the last line. The sum mod 3 should be minus what it currently is. (Since (-1)^x-1 = -1.)

So to establish David's conjecture we must prove that when x is even C(x+1,x-1) - C(x+1,x-4) + ... is a multiple of 3. This is the same as:

C(x+1,2) - C(x+1,5) + C(x+1,8) - C(x+1,11) + ... (*)

because C(n,r) = C(n,n-r).

In fact the expression is a multiple of 3 whether x is even or odd. First note that if x = 0 or 2 (mod 3) then every term in the sum is a multiple of 3. To prove this, write each term in the sum in the form 3p/q C(m,r) where p,q,r are integers with p,q not multiples of 3 and 3m = x+1 or 3m = x (depending on the mod 3 residue of x). Since C(m,r) is an integer it follows each term is a multiple of 3, so this is enough to establish David's conjecture for x = 0 or 2 (mod 3).

If x = 1 (mod 3) write x = 3m + 1. Then note that we have:

C(x+1,2+3r) = C(m,r) (mod 3)

So (*) becomes:

C(m,0) - C(m,1) + C(m,2) - ... = 0

showing that (*) is always 0 (mod 3) which establishes David's conjecture.

There's almost certainly a better way of doing all this but I can't see how. There are probably lots of mistakes/unclear parts here, so please point them out!

I just came across a discussion in the archive that looks related:

Utterly Baffling Infinite Series

Note that we can generalise my first message to find a finite double summation for:

It is:

$\sum _{m=0}^x\sum _{n=0}^{x-m}C(x+1,m)n^x(-1{)}^m{r}^{m+n-x}$ providing of course that the first sum converges.

OK, that's slightly wrong - the r^-x bit should be (1 - 1/r)^x and there are probably more errors in the latest double sum. I will try and post the correct version in the morning when I'm a bit less tired. The double sum in my first message is correct though.

It is possible to do an alternative proof of the divisibility part by noting a curious property of the polynomials p(z) that occur in the numerators. They are all z*q(z) where q(z) is symmetric (in the sense that the leading coefficient is equal to the constant coefficient etc.). The degree of q is odd when x is an even integer, and it follows that if x is odd (1+z) is a factor of q(z); so q(-1) = 0. However we can show quite easily by doing things with congruences that 2^x-1 q(1/2), which is the value of the sum, is congruent to q(-1) mod 3. So we see that the sum is congruent to 0 mod 3 for all even x.
(This is just a very brief sketch; the full proof would take several pages to write out.)

By the way, a slightly weirder conjecture: for large x, it looks rather like

is asymptotic to

$x!/(-\ln z{)}^{x+1}$
Anybody got any idea how on earth we prove this?

David

I have now computed the finite double summation correctly (I hope). In David's notation it is:

By the way David's conjecture is equivalent to showing that for any t then:

where $C$ is a contour enclosing 0 but not $1+2n\pi i$ for any integral $n$.
I don't know if that's going to be helpful either. I'll write the proof of the equivalence in the morning if anyone's interested.

Please do - I presume it has something to do with residues!

Yes, in fact that's a good point. My statement is equivalent to saying that as x-> infinity the residue at 0 of the function t/(z^x (1 - e^(z-t) t)) tends to 1. I still don't see how to prove this...

Here's the proof of the equivalence of David's conjecture, and the contour integral result. We want to show:

Substitute $t=-\ln z$. We want to prove that for $t>0$:

$\sum _{n=1}^{\infty }{n}^x{e}^{-nt}~x!/t^{x+1}$

This is the same as:

$\sum _{n=0}^{\infty }{d}^x/{\mathrm{dt}}^x(e^{-nt})~x!/t^{x+1}(-1{)}^x$

Now if we switch to applied maths mode, we can swap the sum and differential operator.

$d^x/{\mathrm{dt}}^x\sum _{n=0}^{\infty }(e^{-nt})~x!(-1{)}^x/t^{x+1}$

[I think we can justify this step rigorously since each term is monotonic.]

So anyway we desire to prove that:

$d^x/{\mathrm{dt}}^x[1/(1-e^{-t}]~x!(-1{)}^x/t^{x+1}$ (*)

where the derivative on the left hand side is evaluated at $t$. (This is probably an abuse of notation.)

Now Cauchy's integral formula states that if $f$ is analytic within and on a closed contour $C$ then if $z_0$is inside $C$:

$f^{(n)}=n!/(2\pi i){\int }_{C}f(z)\mathrm{dz}/(z-z_0{)}^{n+1}$

(In other words it tells us that if we know everything about an analytic function on the edge of a contour, then we can deduce the value of the function (and its derivatives) anywhere inside the contour.)

Let $f(z)=1/(1-e^{-z})$. We know that the $x$th derivative of $f(z)$ at $z=t$ is given by:

$x!/(2\pi i){\int }_{C}1/((1-e^{-z})(z-t{)}^{x+1})\mathrm{dz}$

So you substitute this back into (*) and replace $z$ with $tz+t$ you get the integral I gave in my last message.

I don't know if this method is leading anywhere, but it seemed like a good idea at the time since the $x!$ cancelled from both sides. I can't seem to find a clever way to incorporate the $x\to \infty$ condition to find the limit of the integral.

Just thought I'd foul everything up a bit:
what's to stop us letting x be non-integral?

I've just realised: this problem boils down to finding the power series expansion of a/(a-e^x ) for integer a. The k^th derivative of this at zero is

I don't actually use the university library very much (it's a long way away and looks like a prison!) but the college library usually has plenty of stuff to keep everyone entertained. Anyway, I'm at home now, but I may drop by Imperial Library in a couple of days, so I'll see if I can find anything out.

Yes, I don't think the contour integral is going to be easier to evaluate in the limit, so your approach of looking directly at the Taylor series of a/(a - e^x ) looks more promising. Along these lines I have "proved" David's statement in totally non-rigorous fashion. I might post the argument later, but it will be for amusement value only!

The proof is quite simple and involves applying the ratio test to the series $\sum _{n=0}^{\infty }{c}_n{z}^n$. The value of $\underset{n\to \infty }{lim}|c_n{z}^n{|}^{1/n}$ is clearly $\left|z\right|\underset{n\to \infty }{lim}|c_n{|}^{1/n}$, so the series converges only when this is $<1$. But we know that the nearest singularity of the function to the origin is at $z=\ln a$, so the radius of convergence must be $\ln a$; comparing this with the formula above we have $\underset{n\to \infty }{lim}|c_n{|}^{1/n}=1/\ln a$ as required.
Is this at all like your "amusement value" proof, Michael, and if so how can it be extended?

Moving on a little, can we prove any useful relations between two functions if we know that the coefficients in their Taylor series are asymptotically equal? My proof above shows that they both have the same radius of convergence, but this is not really very useful.

Incidentally, for "limit" read "limit superior" throughout the above proof.

David

No, my proof isn't anything like that. (Your proof that lim c_n ^1/n = 1/ln a is actually rigorous!!) I have made my "proof" of your original conjecture a little more rigorous by using complex integration but there's still a very dodgy step. At best the following shows why your conjecture is probably true. All pure mathematicians look away now...

Your conjecture is equivalent to:

d^x /dt^x [1/(e^t - 1)] ~ (-1)^x+1 x!/z₀ ^x+1

where the LHS is evaluated at t = z₀ where z₀ > 0.

Using Cauchy's integral formula, we can re-write what we want to prove as:

Now make the substitution $z\text{'}=(x+1)z$:

${\int }_{C*}(e^{z\text{'}/(x+1)}-1{)}^{-1}(x+1{)}^{-1}/(1-z\text{'}/(z_0(x+1{)))}^{x+1}\mathrm{dz}\text{'}~2\pi i$

where $C*$ is the contour $C$ scaled by a factor of $x+1$. But in fact there's no need to distinguish $C*$ and $C$ since they contain the same singularities, and we may as well replace the dummy variable $z\text{'}$ by $z$ so that we want to show:

${\int }_C(e^{z/(x+1)}-1{)}^{-1}(x+1{)}^{-1}/(1-z/(z_0(x+1{)))}^{x+1}\mathrm{dx}~2\pi i$

Now comes the dodgy stage.We know that as $x\to \infty$, $(1-z/(z_0(x+1{)))}^{x+1}\to e^{-z/z_0}$. So it's reasonable to suppose your assertion is equivalent to:

${\int }_C(e^{z/(x+1)}-1{)}^{-1}(x+1{)}^{-1}{e}^{z/z_0}\mathrm{dx}~2\pi i$

or

${\int }_C(e^{z/x}-1{)}^{-1}{x}^{-1}{e}^{z/z_0}\mathrm{dz}~-2\pi i$

The only singularity in this contour is at $z$.

I have found a proper proof, which I think is rigorous (in other words I invite criticism):

Consider $f(z)-g(z)$. As $z$ tends to $\ln a$, this tends to a limit (1/2) (this may be easily proved by expanding them in powers of $(z-\ln a)$.)

Hence the Taylor series of $f(z)-g(z)$, which has coefficients $d_n=c_n-1/(\ln a{)}^{n+1}$, converges at $z=\ln a$. So the $n$th term of this series tends to zero as $n\to \infty$.

But this $n$th term is $d_n(\ln a{)}^n$, so $c_n(\ln a{)}^n-1/(\ln a)\to 0$. Hence $c_n(\ln a{)}^{n+1}\to 1$, so $c_n$ is asymptotically $1/(\ln a{)}^{n+1}$. QED.
Any thoughts?

David

Hmm... simultaneous posts!

Reading your post, I'm confused by one thing.
You seem to have lost the singular points in your integral! The function given - (e^z/x -1)^-1 x^-1 e^z/z₀ - is singular only at 0, which is outside the contour. I think when you let x sort of tend to infinity while keeping the same contour, the singularity receded to infinity and got lost. I presume that is why you abandoned your post halfway through!

Oops lost the rest of my post. It wasn't because I abandoned it though; I'll re-write the last bit tomorrow.

Regarding your proof: I agree that lim(z-> ln a) f(z) - g(z) exists. But how does that show that the Taylor series of f(z) - g(z) converges at z = ln a? Sorry if this is obvious, I'll have another look tomorrow.

because the function h(z)=f(z)-g(z) is analytic in a region containing z=ln a? I'm not sure about this point.

Regarding your post, I think you've ignored uniform convergence. As x increases 1/( 1 - z/(z₀ (x+1) )^x+1 converges pointwise to e^z/z₀ but not uniformly, so its integral does not necessarily converge to the integral of the limit function.

What about 1/(1 + x)? This is certainly analytic in a region containing x = 1, and lim(x-> 1)1/(x+1) exists yet the Taylor expansion doesn't converge at x = 1, and the successive coefficients don't -> 0...

The problem you outline in the second paragraph is precisely the reason it is non-rigorous. I was hoping the fact it wasn't uniformly continuous doesn't matter since it was being multiplied by a function tending to 0. I really have no idea how to justify that stage.

Sorry, in the first paragraph I meant to say the successive terms don't -> 0 in the Taylor series of 1/(1 + x) at x = 1.

Aha! So (just to check) is the following guarenteed?

If a function f(z) is analytic within a circle centre a, radius r then f(z) can be written as a Taylor series about z = a, and the series is guarenteed to converge at each point inside the circle.

I did hear something to this effect, but I didn't know the exact condition.

If this is right then that's an extremely clever proof. What made you think of it?

Regarding my original attempt, it is not too hard to finish it off from there. Substitute t = e^-z/(x+1) which puts the singularity at the origin and makes the contour finite. Then take (wlog) z₀ to be a multiple of x+1, expand the function about the origin and you obtain the answer upon applying the residue theorem. But you're right, the step where I take the limit is not at all rigorous and all my attempts to justify the limit swapping have failed, so it's better to stick to your proof.

Actually the more I think about it the more ridiculous my whole "proof" seems. Effectively I have replaced z with z/(x+1) then replaced this with z again, taking the opportunity (in the intermediate stage) to swap two limits. Effectively I have replaced (1 - x)ⁿ by e^-nx for large x, which while being of the correct order of magnitude, can't easily be justified. Let's just forget that posting :)

Yes, that is the condition for the convergence of Taylor series. It's a wonderful theorem and provides all sorts of neat results, and allows you to write down immediately the radius of convergence of any Taylor series of an analytic function (and non-analytic functions don't have sensible taylor series at all!)

As for how I thought of my proof, it was more or less the obvious extension of my comment about the relationship between functions when they have asymptotic Taylor coefficients: I plotted 2/(2-e^x ) and 1/(ln(2)-x) and noticed that they approached each other as they neared the singularity at ln 2.

By the way, I still don't see how your integral can avoid being zero. But I've forgotten about it already ;-)

David

Would it be possible to evaluate David's above sum (of sqrt(n)/2ⁿ ) using fractional calculus?

We know that