Search Results

Here is a (surprising) proof using Cauchy-Schwarz and "rearrangement". The following lemma will be the key. Lemma : Let $X,Y$ be independent integer-valued rvs, then \begin{align*} (a)\; &\mbox{ for …

Here is another approach, which also gives the rational term. (I) To see how it works let $n\geq 2$ and consider first the simpler case \begin{align*} \mathbb{E}\bigg(\frac{1}{X_1+\ldots+X_n}\bigg)= …

Here is a proof which also shows the exponential decay in $n$ for $p\neq q$ Let $0<p,q<1$ and $B_{n,p},B_{n,q}$ independent $\mathrm{Binomial}(n,p)$ resp. $\mathrm{Binomial}(n,q)$ distributed random v …

Here's an alternative proof based on probabilistic arguments (showing different aspects). Let $$f_n(x):=\sum_{j=0}^n { x \choose j}=[t^n]\,\frac{(1+t)^x}{1-t}\;\;,$$ and let $^\prime$ denote deriv …

Here is a proof which doesn't use the identity $\int_{-\infty}^\infty {n \choose x}\,dx= 2^n$: Using the representation ${ n \choose x}=\frac{1}{2\pi}\int_{-\pi}^\pi e^{-ixt}\left(1+e^{it}\right)^n\, …

One can also use the binomial transform. (If $A(z)=\sum_{i\geq 0} a_i z^i$ is a (formal) power series, the (formal) power series $B(z):=\frac{1}{1-z} A(\frac{z}{1-z})$ has coefficients $[z^n] B(z)=\su …

I sketch the arguments for $C(x)$, the arguments for $L(x)$ are essentially the same. The specific form of the sum suggests probabilistic arguments. Let $X_x$ be a $\mathrm{Poiss}(x^2)$-distributed …