TITLE: A limit involving binomial coefficients: $\lim_{n\to\infty} (-1)^n\sum_{k=1}^n(-1)^k{n\choose k}^{-1/k}=\frac12$?
QUESTION [39 upvotes]: 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 $\lambda(x)=x(1-x)^{(1-x)/x}$.

REPLY [8 votes]: 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.