0
votes
0answers
24 views

Approximation of sum with binomial summands

I am new here, so hopefully my question will be understood correctly. I have a function (originating from expected untility theory in economics) that looks the following: ...
2
votes
2answers
120 views

Sums of central binomial coefficients

Are there closed forms for $$\sum^n_{i=0} \binom{2i}{i}$$ and $$\sum^n_{i=0} \binom{2i}{i}^2$$? Also, how can these sums be approximated?
2
votes
0answers
86 views

Good upper bound on a binomial sum

What is a good upper bound on the following binomial sum: $$\sum_{i,j< \frac{m}{n}} {m \choose i}{m-i \choose j} z^{i+j}$$ where $z = \frac{1}{n-2}$?
1
vote
1answer
75 views

Weighted sum of ratio of partial sum of binomial coefficients

I would like to approximate the following sum when $n \rightarrow \infty$ and $n \gg k$, $$\sum_{x = k}^n \sum_{y > x}^n \frac{\sum_{m = 0}^{k - 1} {y - 2 \choose m}}{\sum_{m = 0}^k {y - 1 \choose ...
4
votes
1answer
79 views

How to calculate (or approximate) “trimmed” (a+b)^n?

$a^n + C_n^{1}a^{n-1}b + ... C_n^{n-1}a^{1}b^{n-1}+b^n = (a+b)^n$ But how to calculate (maybe approximately) $a^n + C_n^{1}a^{n-1}b + ... C_n^{i}a^{n-i}b^{i} = ?$ For info, the underlying problem ...
1
vote
0answers
46 views

Approximation of factorial - Stirling formula [duplicate]

Possible Duplicate: Elementary central binomial coefficient estimates How can I prove that $$ \binom{n}{n/2} = \Theta\left(\frac{2^n}{\sqrt n}\right) $$ I tried with Stirlings ...
1
vote
1answer
322 views

Using binomial theorem find general formula for the coefficients

Using binomial thaorem (http://en.wikipedia.org/wiki/Binomial_theorem) find the general formula for the coefficients of the expantion: $$ ...
3
votes
1answer
86 views

is the approximation of the sum true?

Someone commented under my question Calculation of the moments using Hypergeometric distribution that $$ \sum_{k=0}^l\frac{{l \choose k}{2n-l \choose n-k}(2k-l)^q}{{2n\choose n}}\sim \sum_{k=0}^l ...
3
votes
1answer
226 views

Surprising approximation of weighted sum of binomial coefficients

The following sum appeared in connection to the problem addition of angular momentum in physics: $$ \frac{1}{2^{n+3}}\sum_{k=0}^n \left(\frac{n-2k-1}{\sqrt{k+1}}+\frac{n-2k+1}{\sqrt{n-k+1}}\right)^2 ...
0
votes
0answers
110 views

Calculation of sum

I am wondering if it is possible to calculate or approximate the following sum $$ \sum_{k=0}^l\frac{(l-2k)^p(2l+k(k-1))l^{k-1}}{(k+3)(k+2)} $$here $p \geq 2$. Thank you.
6
votes
2answers
394 views

Elementary bound of binomial coefficient

I'm working my way through an Erdős paper from the sixties and some of the elementary bounds he claims seem to be just beyond my reach. The expression looks horrendous but maybe there is a clever ...
1
vote
0answers
282 views

Calculation of a 'double' sum

Let $n \in N$ and $q\geq 2$. I am trying to calculate the following sum: $$ \sum_{i=0}^{\sqrt n/2}\sum_{j= i \sqrt n }^{(i+1)\sqrt n}\frac{(-1)^q2^q(\frac{n}{2}-j)^q}{(n-j)!j!} $$ Any help will be ...
4
votes
1answer
191 views

Evaluating a limit involving binomial coefficients.

If $N_c=\lfloor \frac{1}{2}n\log n+cn\rfloor$ for some integer $n$ and real constant $c$, then how would one go about showing the following identity where $k$ is a fixed integer: $$\lim_{n\rightarrow ...
4
votes
2answers
373 views

computation of the sum

I am having trouble to compute the following sum: $$ \sum_{k=0}^n(n-2k)^p \frac{{n \choose k}{2m-n \choose m-k}}{{2m \choose m}} $$ Here $p\geq 2$. To simplify the question, we can even assume that ...
9
votes
2answers
3k views

Approximating the logarithm of the binomial coefficient

We know that by using Stirling approximation: $\log n! \approx n \log n$ So how to approximate $\log {m \choose n}$?
5
votes
3answers
3k views

How to simplify or calculate a formula with very big factorials

I'm facing a practical problem where I've calculated a formula that, with the help of some programming, can bring me to my final answer. However, the numbers involved are so big that it takes ages to ...