Show that $ 2^n\sum_{N=1}^{[R]}\binom{N+n-1}{n-1}\le (\frac{CR}{n})^n $.

Question

Recall that Stirling's formula states that for any integers $k\ge 1$, we have $$ k!=(1+o(1))\sqrt{2\pi k}(\frac{k}{e})^k. $$

I try to show that for integers $n\ge 1, N\ge 1$ and $R>n$, we have $$ 2^n\sum_{N=1}^{[R]}\binom{N+n-1}{n-1}\le (\frac{CR}{n})^n $$ for some positive constant $C>0$ which is independent on $R$ or $n$.

I try to apply the Stirling's formula: $$ \sum_{N=1}^{[R]}\binom{N+n-1}{n-1}=\sum_{N=1}^{[R]}\frac{(1+o(1))}{(1+o(1))^2\sqrt{2\pi}}\frac{\sqrt{N+n-1}(N+n-1)^{N+n-1}}{\sqrt{N(n-1)}(n-1)^{n-1}N^N} $$

I am stuck here... How to upper this summation?

$$\sum_{N=0}^r\binom{N+m}m=\binom{m+r+1}{m+1}$$ Letting $m=n-1,$ $r=[R],$ your sum is $\binom{[R]+n}{n}-1.$ — Thomas Andrews, May 06 '24 at 23:01
This doesn't look right, because, as $n\to\infty,$ the right side goes to zero while the left side goes to $+\infty.$ And if $C$ is supposed to depend on $n,$ then why not just replace the whole right side with $DR^n$ for some $D$ depending on $n.$ — Thomas Andrews, May 06 '24 at 23:10
Is it maybe supposed to be $2^{-n}?$ That doesn't look right, either, but may be closer. — Thomas Andrews, May 06 '24 at 23:12
@ThomasAndrews Sorry, we need to add extra condition that is $R>n$. — Hermi, May 08 '24 at 16:32
Add it to the question. A person shouldn't have to read through all the commenta to know what your question means. — Thomas Andrews, May 08 '24 at 18:13
@ThomasAndrews Absolutely! I already added it in my question. — Hermi, May 08 '24 at 22:48

Amir · Accepted Answer · 2024-05-15T07:03:10.827

We know that [1]:

$$\left(\frac{n}{k}\right)^k \leq {n \choose k} \leq \left( \frac{en}{k}\right)^k$$

Hence, using the identity [2] also suggested by @ThomasAndrews in a comment, for $[R]>n$ we have $$\sum_{N=1}^{[R]}\binom{N+n-1}{n-1}=\binom{[R]+n}{n}-1\le\binom{2[R]}{n} \le \\ \left(\frac{2e[R]}{n}\right)^n\le \left(\frac{2eR}{n}\right)^n,$$

which yields the result for $C=4e$ after multiplying both sides by $2^n$.

Show that $ 2^n\sum_{N=1}^{[R]}\binom{N+n-1}{n-1}\le (\frac{CR}{n})^n $.

1 Answers1