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Experimentation suggests the limit $$\lim_{n\rightarrow\infty} (-1)^n\sum_{k=1}^n(-1)^k{n\choose k}^{-1/k}=\frac{1}{2}\ .$$ Does somebody have an idea for (a start of) a proof?

Added: There seem to be variations: $$\lim_{n\rightarrow\infty} (-1)^n\sum_{k=1}^n(-1)^k{2n\choose k}^{-1/k}=\frac{1}{8}\ ,$$ $$\lim_{n\rightarrow\infty} (-1)^n\sum_{k=1}^n(-1)^k{3n\choose k}^{-1/k}=\frac{2}{27}\ ,$$ etc.

Moreover, the exponent $-1/k$ in the original identity can seemingly be replaced by $-2/k$, $-3/k$ (by $-\alpha/k$ for strictly positive $\alpha$?) without changing the limit value.

Update: Given a strictly positive rational number $\frac{p}{q}\leq 1$ (with coprime natural integers $p,q$), there is perhaps a number $\lambda(p/q)$ such that $$\lim_{n\rightarrow\infty}(-1)^n\sum_{k=1}^{pn}(-1)^k{qn\choose k}^{-\alpha/k}= \frac{(\lambda(p/q))^\alpha}{2}$$ for $\alpha$ real and strictly positive. A few values for $\lambda$ are seemingly $$\lambda\left(\frac{1}{3}\right)=\frac{4}{27},\ \lambda\left(\frac{1}{2}\right)=\frac{1}{4},\ \lambda\left(\frac{2}{3}\right)=\frac{2}{3\sqrt{3}},\ \lambda\left(1\right)=1\ .$$ Two more conjectures: $\lambda\left(\frac{1}{q}\right)=\frac{(q-1)^{q-1}}{q^q}$ and $\lambda(x)^x=\lambda(1-x)^{1-x}$ for rational $x$ in $(0,1)$.

The correct formula for $\lambda$ is perhaps $\lambda(x)=x(1-x)^{(1-x)/x}$.

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What led you to conjecture this? It's very unusual-seeming (to me, anyhow). –  Igor Rivin Feb 2 '12 at 18:56
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Serendipity (playing around with Maple). –  Roland Bacher Feb 2 '12 at 19:04
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What happens if you use Stirlings Formula to estimate the factorials in the binomial coefficient? A back-of-an-envelope calculation seems to indicate some nice cancellation... –  Kevin Buzzard Feb 2 '12 at 19:14
    
@Kevin Buzzard: I agree but how can one exploit this! –  Roland Bacher Feb 2 '12 at 19:20
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You would need to have (for $n$ even), $\sum_{k=pn+1}^{k=pn+n} (-1)^k \left(\begin{array}{c} qn \\ k \right)^{-1/k}=\lambda(p+1/q)-\lambda(p/q)$. If $qn$ has a lot of prime factors, this seems to imply that a very short chunk should look exactly like the derivative of $\lambda$. Is the difference of two terms propoprtional to $d\lambda$? –  Will Sawin Feb 4 '12 at 22:03

3 Answers 3

Off-the-wall suggestion... Take $n$ even, I call it $2n$ now. Then asymptotically as $n \to \infty$ $$ \binom{2n}{2n-2j-1}^{-1/(2n-2j-1)} - \binom{2n}{2n-2j}^{-1/(2n-2j)} \sim \frac{1}{2n}\log \frac{2n}{2j-1} $$ and the sum $$ \frac{1}{2n}\sum_{j=1}^{n}\log\frac{2n}{2j-1} $$ is a Riemann sum for the integral $$ \frac{1}{2}\int_0^1 \log\frac{1}{t}\;dt = \frac{1}{2} . $$

Added Feb.3
I said it was off-the-wall. The asymptotic expansion is from Maple, like this:

asymptotics

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Sounds pretty reasonable. I assume all you need to do from here is bound the error... –  Igor Rivin Feb 2 '12 at 20:17
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I am not sure that your asymptotic expansion is correct (it seems not to work numerically, for $n=100000,j=50000$, the error is $\sim 1/144267$ which is of order $0(1/n)$). How do you get it? –  Roland Bacher Feb 3 '12 at 9:52
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Your method should also apply to $(-1)^n\sum_{k=1}^n(-1)^k{2n\choose k}^{1/k}$ which converges numerically quite convincingly to $1/8$ while your method suggests $(1+\log 2)/4$ as the limit. –  Roland Bacher Feb 3 '12 at 10:19
    
@Gerald Edgar: Sorry, I did misunderstand the expression "off-the-wall". I guess the asymptotics is correct for fixed $j$ and $n$ going to infinity. The correct value of the integral is thus probably a happy coincidence, since contributions to the limit come from most terms of the sum (except the very first which can of course be neglected). –  Roland Bacher Feb 3 '12 at 15:58
    
@Gerald: Roland seems to be correct. You need the bulk of the terms, but your asympt() is for fixed $j$. You also have the sign wrong. Putting $j=an$ before applying asympt(), assuming $0<a<1$, gives for your first expression $a^{a/(1-a)}\ln(a)/(2(1-a)n)$, which integrates to exactly $-1/(2n)$. So it does seem like a happy coincidence. The same approach also verifies that the answer (remarkably) doesn't change if the exponent $-1/k$ is replaced by $-c/k$ for $c\ge 1$. –  Brendan McKay Feb 6 '12 at 21:57

I would suggest to re-write the problem as $$\lim_{n\to\infty} \sum_{k=0}^{n-1} (-1)^k \binom{n}{k}^{-1/(n-k)} = {}?\,.$$ Now if I'm not mistaken, it is true that for fixed $x$ with $|x| < 1$, we have $$\lim_{n\to\infty} \sum_{k=0}^{n-1} (-1)^k \binom{n}{k}^{-1/(n-k)} x^k = \frac{1}{1+x}$$(the coefficients have absolute value $\le 1$ and converge termwise to $(-1)^k$; multiplying by $x^k$ lets us apply dominated convergence). So it would be sufficient to show that $\lim_{n \to \infty}$ and $\lim_{x \to 1-}$ commute. I don't have time right now to really dive into it, but maybe somebody else would like to go from here.

Note that this would also give a proof of the more general version: $$\lim_{n\to\infty} \binom{qn}{pn+k}^{-a/(n-k)} = \lim_{n\to\infty} \binom{qn}{pn}^{-a/n} = \left(\frac{p^p (q-p)^{q-p}}{q^q}\right)^a$$

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Here is a different approach. Write $$a_{n,k} = \binom{n}{k}^{-1/(n-k)}$$ (or whatever the term is) for $0 \le k < n$ and $a_{n,k} = 0$ for $k \ge n$. Then $$\sum_{k=0}^{n-1} (-1)^k \binom{n}{k}^{-1/(n-k)} = \sum_{k=0}^\infty (-1)^k a_{n,k} = \frac{a_{n,0}}{2} + \frac{1}{2} \sum_{k=0}^\infty (a_{n,2k} - 2 a_{n,2k+1} + a_{n,2k+2})\,.$$ So if you can show that $$\lim_{n\to\infty} \sum_{k=0}^\infty |a_{n,2k} - 2 a_{n,2k+1} + a_{n,2k+2}| = 0\,,$$ the claim will follow. This should hold for any sufficiently "smooth" double sequence $(a_{n,k})$, since it just means that locally the sequence (for fixed $n$) varies nearly linearly.

If $a_{n,k} = f_n(k)$ for some nice function $f_n$, then the second order difference is bounded by $\max \{|f''_n(x)| : 2k \le x \le 2k+2\}$, which may help in the estimate. For example, taking $a_{n,k} = n^{-1/(n-k)}$, so $f_n(x) = \exp\left(-\frac{1}{n-x} \log n\right)$, we have $$f''_n(x) = \left(\left(\frac{\log n}{(n-x)^2}\right)^2 + \frac{2\log n}{(n-x)^3}\right) f_n(x)\,.$$ For $0 < \alpha < 1$, we find that $$\sum_{k=0}^{\lfloor n - \alpha \log n\rfloor} |a_{n,2k} - 2 a_{n,2k+1} + a_{n,2k+2}| \le C \frac{1}{\alpha^3 \log n}\,,$$and the tail can be estimated by a constant times $\exp(-1/\alpha)$. Taking $\alpha$ arbitrarily close to zero shows that $$\lim_{n\to\infty} \sum_{k=0}^{n-1} (-1)^k n^{-1/(n-k)} = \frac{1}{2}\,.$$ For the original problem, the estimates are likely to be more involved, but should be possible.

Addition:
The formula for $\sum_{k=0}^n (-1)^k a_{n,k}$ given near the beginning of this answer can (in the case $a_{n,k} = f_n(k)$) be interpreted as $a_{n,0}/2$ plus half the difference between the trapezoid-rule and midpoint-rule approximations to $\int_0^\infty f_n(x)\,dx$ using the even integers as subdivision points. For a well-behaved sequence of functions, this difference should tend to zero.

So the problem has only little to do with binomial coefficients as such.

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