# Tag Info

17


3

I was also working with the inequality $$\left(1 - \frac{k}{n}\right)^{n-k} \leq \exp \left( -k + \tfrac{k^{2}}{n} \right) = \exp\left\{ -n \cdot \tfrac{k}{n} \left( 1 - \tfrac{k}{n} \right) \right\}.$$ Now note that $q(x) = x(1-x)$ satisfies $q(1-x) = q(x)$ and $q(x) \geq \frac{1}{2}x$ on $[0, \frac{1}{2}]$. Then we get $$0 \leq k \leq \tfrac{1}{2}n ... 3 To get an idea of the asymptotics, try using Stirling's approximation on the binomial coefficients:$$\ln {n \choose k} \approx n \ln n - k \ln k - (n - k) \ln(n - k).$$Combining this with the other term in the summation,$$\ln \left(\frac{c k}{n}\right)^k = k \ln c + k \ln k - k \ln n,$$we get$$\ln \left[{n \choose k} \left(\frac{c ...

3

Clearly $$\frac{1}{(k-1)k(k+1)}=\frac{1}{2}\left(\frac{1}{k(k-1)}-\frac{1}{k(k+1)}\right).$$ Thus $$4\sum_{k=1}^{n-1} \frac{1}{k(k+1)(k+2)}=2\sum_{k=1}^{n-1}\left(\frac{1}{k(k+1)}-\frac{1}{(k+1)(k+2)}\right)=\frac{2}{1\cdot 2}-\frac{2}{n(n+1)}.$$ Thus $$\lim_{n\to\infty}4\sum_{k=1}^{n-1} \frac{1}{k(k+1)(k+2)}=1.$$

2

Expand by partial fractions: $$\frac{1}{(n - 1)n(n + 1)} = -\frac 1 n + \frac 1 2 \frac 1 {n - 1} + \frac 1 2 \frac 1 {n + 1}$$ Now notice that a lot of the terms will cancel, and the series telescopes. This is similar to, and motivated by the fact that $$\sum_{n = 1}^{\infty} \frac 1 {n(n + 1)} = \sum_{n = 1}^{\infty} \frac 1 n - \frac 1 {n + 1}$$ ...

2

Suppose for the sake of contradiction that there is no real positive $n$ such that $$(n+1) \left (1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}\cdots\dfrac{1}{n+1} \right ) \geq 5280n.$$ Then we would have, for all $n$, $$\left (1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}\cdots\dfrac{1}{n+1} \right ) < 5280\frac{n}{n+1}<5280,$$ implying that the infinite ...

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