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8
votes
7answers
3k views

Lower bound for sum of binomial coefficients?

Hi! I'm new here. It would be awesome if someone knows a good answer. Is there a good lower bound for the tail of sums of binomial coefficients? I'm particularly interested in the simplest case ...
2
votes
3answers
996 views

Estimating a partial sum of weighted binomial coefficients

There is a well-known estimate for the sum of all binomial coefficients $\binom{n}{k}$ satisfying $k \leq \alpha n$ for some $\alpha$ satisfying $0 < \alpha \leq 1/2$: $$ \sum_{k=0}^{\alpha ...
31
votes
4answers
2k views

Integer-valued factorial ratios

This historical question recalls Pafnuty Chebyshev's estimates for the prime distribution function. In his derivation Chebyshev used the factorial ratio sequence $$ ...
36
votes
2answers
2k views

Alternating sum of square roots of binomial coefficients

Let $$ c_n = \sum_{r=0}^n (-1)^r \sqrt{\binom{n}{r}}. $$ It is clear that $c_n = 0$ if $n$ is odd. Remarkably, it appears that despite the huge positive and negative contributions in the sum ...
17
votes
3answers
2k views

Need help proving that $\sum\limits_{j=0}^{k-1}(-1)^{j+1}(k-j)^{2k-2} \binom{2k+1}{j} \ge 0$

Hello. I have been trying very hard to show that $\sum\limits_{j=0}^{k-1}(-1)^{j+1}(k-j)^{2k-2} \binom{2k+1}{j} \ge 0$ and could not quite get anywhere. This inequality has been verified by computer ...
8
votes
2answers
811 views

Proving $\sum_{k=0}^{2m}(-1)^k{\binom{2m}{k}}^3=(-1)^m\binom{2m}{m}\binom{3m}{m}$

I found the following formula in a book without any proof: $$\sum_{k=0}^{2m}(-1)^k{\binom{2m}{k}}^3=(-1)^m\binom{2m}{m}\binom{3m}{m}.$$ This does not seem to follow immediately from the basic ...
2
votes
0answers
133 views

Sum of binomial coefficients weighted by a lower incomplete regularized gamma function

The following summation turned up in the course of my research: $$S_n=\sum_{k=0}^n {n \choose k}\lambda^k P(k,t)$$ where $P(k,t)=\frac{1}{\Gamma(k)}\int_{0}^t e^{-x}x^{k-1}dx$ is the lower ...
6
votes
3answers
442 views

Is there a closed formula for the generating function of some trinomial coefficients?

We learn in calculus how to obtain a sum of binomial coefficients \frac{(2d)!}{(d!)^2} in terms of a generating function $\sum_{d \geq 0} \frac{(2d)!}{(d!)^2} x^d$ by the Taylor series of ...
0
votes
1answer
718 views

How to calculate such sum of product of binomial coefficient?

..I wonder if the following formula can be calculated? $ \sum_{k=0}^m {m \choose k} {2k \choose n} $