How to prove this series converges to an irrational limit?

Question

I think this is interesting:

Let k >= 2 be an integer and let p be a polynomial of degree >= 2, with integer coefficients, such that the coefficient of the leading term is
positive. Show that the series Sum(n = 1, oo) 1/[k^(p(n)] converges to an irrational number.

OK, it seems there won’t be any more answers. I’ll give the proof I came up with. In base k, the number k is represented by 10. For each integer n, p(n) is an integer. Since the leading term of p is positive, for n sufficient large p(n) and p(n) – p(n-1) are positive and both goes to oo as n à oo. Choose m such that p(n) and p(n) – p(n-1) are positive for n >m. Then, the limit L is given by the sum of N = k/p(1) …+ k/p(m-1 and W = Summation (n = m, oo) k/p(n). L = N +W, with N rational. Now, let’s show W is irrational.

For n >m, the representation of k/p(n) in base k is 0.000…..1, with p(n)- 1 0’s before 1. p(n) – p(n-1) is the “distance” between the position of 1 in 2 consecutive terms of the series. Since this “distance” goes to oo, the representation of W in basis k can be neither finite nor infinite and periodic. So, W is irrational. And since L = N + W, L is irrational.

Answer 1

This was tough, but I think I've gotten it.

Let's consider the special case which just barely meets your criteria:
A = Summation (n=1, infinity)[1/k^(n^2)]
We want to show that the assumption that
A = p/q for two integers p and q results in a contradiction.

Define:
S(N) = Summation (n=1, N)[1/k^(n^2)]
Then:
A = S(N) + Summation (n=N+1, infinity)[1/k^(n^2)]
So:
A - S(N) = Summation (n=N+1, infinity)[1/k^(n^2)]
= (1/k^(N+1)^2 + 1/k^(N+2)^2 + 1/k^(N+3)^2 +..)
= (1/k^(N^2))*(1/k^(2N+1) + 1/k^(4N+4) + 1/k^(6N+9)+...)
= (1/k^(N^2))*((1/k^(2N) /k + (1/k^(2N))^2 /k^4
(1/k^(2N))^3/k^9 +..)

Defining
x = (1/k^(2N)), this is:
A - S(N) = (1/k^(N^2))*(x/k + x^2/k^4 + x^3/k^9 +...)
= (x/k^(N^2))*(1/k)*[1 + x^1/k^3 + x^2/k^8 + ...]
We can drastically overestimate the sum in square brackets by:
[1 + ..] < 1 + x + x^2 + ...
= 1/(1-x)

Therefore,
A - S(N) < (x/(1-x))*(1/k)*(1/k^(N^2))
= (1/(1/x - 1))*(1/k)*(1/k^(N^2))
= (1/(k^(2N) - 1) * (1/k) * (1/k^(N^2))
= k^(-(N^2+1))/(k^(2N) - 1)

OK, we're almost there:
To recap,
A = S(N) + epsilon , where
epsilon = A - S(N)
< k^(-(N^2+1))/(k^(2N) - 1)

(The important thing to notice is that we can make epsilon very tiny. Watch!)

Since A = p/q
p/q = S(N) + epsilon

We now multiply both sides by: q*k^(N^2):
p*k^(N^2) = q * k^(N^2) * S(N) + epsilon * q * k^(N^2)

The left-hand-side: obviously a product of integers is an integer.
The first term on the right-hand-side: every term in the summation S(N) is of the form (1/k^(integer)), with the integer less than or equal to N^2. Therefore, the first term on the RHS is a sum of integers and is thus an integer.
Therefore, the remainder must also be integer. But how big is it?
epsilon*q*k^(N^2) < k^(-(N^2+1))/(k^(2N) - 1)*q*k^(N^2)
= q * k^(-1)/(k^(2N) - 1)
= q/(k*(k^(2N) - 1)

What this says is that, once you have told me what q is, I can easily find a value of N so large that the error (A - S(N)) is too small to make up the difference required to keep everything integer-valued.

In other words, either A is not p/q, or else A - S(N) = 0 already. But the second cannot be true, because we are still adding positive terms to the sequence beyond n = N. So it cannot be true that the sum is rational.
QED.

OK, that proves it for the simplest case:
p(n) = n^2
How can we generalize it?
a) p(n) = a*n^2 : piece of cake, change k to k^(a), everything goes through unchanged.
b) p(n) = n^2 + b*n + c : In the calculation of epsilon, we can use the same over-estimate. It's already so much overkill that this doesn't make any difference. The rest of the argument goes through unchanged.
c) p(n) = a*n^2 + b*n + c: Do a) and then b).
d) p(n) = n^3 : Play the same game: Over-estimate A - S(N), and it will still be less than something like:
k^(-N^3)/(k^(3N^2) - 1). Even when you multiply by q*k^(N^3), there's no chance of keeping this an integer. The same argument works.
e) It's now obvious that p(n) can be generalized to any case you mentioned: powers higher than n = 2, with positive leading coefficient. The over-estimation strategy keeps on working.

P.S. I guess I forgot to show above that the summation actually converges. Well, just compare the terms:
1/k^(n^2) to 1/k^(n) : the right-hand terms clearly dominate the left-hand terms, and the right-hand terms clearly define the geometric series. So by comparison, the first one converges as well. The same proof applies, of course, to all the generalizations above.

Answer 2

I think that the sum is irrational. Let,

a/b = 1/k^p(1) + 1/k^p(2) + ...+1/k^p(n) + 1/k^p(n+1) +...

for integers a and b. Multiply both sides by bk^p(n) and

look at the size of S = b/(k^(p(n+1)-p(n)) + ... where the

difference p(n+1)-p(n) exceeds b for n>c when n is chosen large enough.

This forces S<1 and we have the necessary contradiction.

Answer 3

this one is pretty vicious, but if i were to tackle it i'd use an induction argument

first you show that for the series is irrational (i think these work themselves out to something related to logarithms, i don't remember though, i'd recommend checking a table of famous series identities)

next your gonna wanna go for this n+1 step, intuitively i feel that since you know that the sum from n=1..inifinity works when induct on n and n=1..infinity+1 you might have to make another limit argument or maybe, you will have something like an irrational * some number (provided the number is not the reciprocal of the irrational) is irrational (ex pi*3 is 3pi is irrational)

basically if you get the first part you got the second part, good luck (wikipedia and mathworld are good websites to explore for this stuff)

Answer 4

i do no longer think that's actual. evaluate, as an occasion, if bn = a million/r^n. Then bn >= 0 for all n, yet b1r^a million = a million, b2r^2 = a million, and for each n, bnr^n = a million. because of the fact the shrink of bnr^n isn't 0, the series can not converge.