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How can I show that $ \sum_{k=0}^{m}\binom{n}{k} \leq (n+1)^m$ for $m \leq n $?

mathpete
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1 Answers1

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$(n+1)^m =\sum_{k=0}^m \binom{m}{k}n^k $ so if $n^k\binom{m}{k} \ge \binom{n}{k} $ we are done.

$\begin{array}\\ \binom{m}{k} &=\dfrac{m!}{k!(m-k)!}\\ &=\dfrac{\prod_{j=0}^{k-1}(m-j)}{k!}\\ \dfrac{n^k\binom{m}{k}}{\binom{n}{k}} &=\dfrac{n^k\prod_{j=0}^{k-1}(m-j)}{\prod_{j=0}^{k-1}(n-j)}\\ &\ge\prod_{j=0}^{k-1}(m-j) \qquad\text{since } n^k \ge \prod_{j=0}^{k-1}(n-j)\\ &\ge 1\\ \end{array} $

marty cohen
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    I do not know why this question has been closed since the summation is up to $m$ and not $n$. Interested by https://math.stackexchange.com/questions/3912163/sum-of-binomial-coefficients-so-that-the-sum-equals-n-choose-n-2/3912290#3912290 I wonder if you are aware fo a "simple" upperbound in order to possibly improve my estimate or at least derive the asymptotics. Thank you. – Claude Leibovici Nov 27 '20 at 07:04
  • I agree. I find it surprising the number of questions which are incorrectly closed. So I voted to reopen. At least it turns out that there is a simple answer. – marty cohen Nov 27 '20 at 11:37
  • I found a very interesting approximation in https://math.stackexchange.com/questions/103280/asymptotics-for-a-partial-sum-of-binomial-coefficients which allows a good approximation (if you have time to waste, have a look at my last update in https://math.stackexchange.com/questions/3912163/sum-of-binomial-coefficients-so-that-the-sum-equals-n-choose-n-2/3912290#3912290 ) – Claude Leibovici Nov 28 '20 at 09:58