How did the following equation follow?
$$\ln1+\ln2+\dots+\ln k\ge \int_{x=1}^k\ln x\mathop{dx}=k\ln k-k+1$$
I need to show $\binom{n}{k}<\frac{1}{e}\left(\frac{en}{k}\right)^k$.
But we have $\binom{n}{k}<\frac{n^k}{k!}$.
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1Can you explain more your issue? Do you have issue with proving the first line inequation? Do you have issue using this inequation to prove the second inequation. What about the third inequation? Why does it stands here? – Martigan Jan 06 '22 at 17:27
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See my answer to this question, for example. – Greg Martin Jan 06 '22 at 17:27
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Break the integral into $k$ pieces, over the intervals $[j.j+1]$ (for $1\leq j<k$) and note that, on $[j.j+1]$, the integrand $\ln x$ is $\geq\ln j$. – Andreas Blass Jan 06 '22 at 17:28
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Can you prove/argue why $\ln m \ge \int_{x=m-1}^m \ln x\ dx$? (Hint: $\ln x$ is monotonically increasing so for any $x \in [m-1, m]$ we have $\ln x \le \ln m$ so $\int_{x=m-1} \ln x\ dx \le \int_{x=m-1}^m \ln m\ dx$. Hint 2: $\ln m$ is a constant) – fleablood Jan 06 '22 at 17:43
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Notice that $\ln n=\int_{n-1}^n\ln n\mathop{dx}$ for all $n=2,3,\dots, k$. Therefore,
\begin{align*} \ln 1+\ln 2+\ln 3+\dots+\ln k&=0+\int_{1}^2\ln 2\mathop{dx}+\int_{2}^3\ln 3\mathop{dx}+\dots+\int_{k-1}^k\ln k\mathop{dx}\\ &\ge \int_{1}^2\ln x\mathop{dx}+\int_{2}^3\ln x\mathop{dx}+\dots+\int_{k-1}^k\ln x\mathop{dx}\\ &=\int_{1}^k\ln x\mathop{dx}. \end{align*}
This inequality occurs since $\ln x$ is a strictly increasing function.
Bonnaduck
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