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Binomial coefficients as multiple sum

Prove that

    \[\sum_{n_1=1}^{n-1} \sum_{n_2=1}^{n_1-1} \sum_{n_3=1}^{n_2-1} \cdots \sum_{n_m=1}^{n_{m-1}-1} 1 = \binom{n-1}{m}\]

Solution

The binomial coefficient in the RHS enumerates the subsets A of size m of \{1,2,\ldots,n-1\}. The LHS does the same thing, but choosing first the largest element n_1 of A, then its second-to-largest element n_2 <  n_1, until choosing its smallest element n_m.

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On the factorial

Let \mu denote the Möbius function and \left \lfloor \cdot \right \rfloor denote the floor function. Prove that:

    \[n! = \prod_{j=1}^{\infty} \prod_{i=1}^{\infty} \left ( \left \lfloor \frac{n}{ij} \right \rfloor! \right )^{\mu(i)}\]

Solution

The RHS equals

    \begin{align*} \prod_{j=1}^{\infty} \prod_{i=1}^{\infty} \left ( \left \lfloor \frac{n}{ij} \right \rfloor! \right )^{\mu(i)} &= \prod_{k=1}^n \prod_{i|k} \left( \left \lfloor \frac{n}{k}\right\rfloor!\right)^{\mu(i)} \\ &= \prod_{k=1}^n \left( \left \lfloor \frac{n}{k}\right\rfloor!\right)^{\sum_{i|k}\mu(i)} \\ &= n! \end{align*}

since \sum \limits_{i | k} \mu(i) = 0 for k>1.

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Sum over all positive rationals

For a rational number x that equals \frac{a}{b} in lowest terms , let f(x)=ab. Prove that:

    \[\sum_{x \in \mathbb{Q}^+} \frac{1}{f^2(x)} = \frac{5}{2}\]

Solution

First of all we note that

    \[F(s) = \sum_{x \in \mathbb{Q}^+} \frac{1}{f^s(x)} = \sum_{\substack{a,b=1 \\ \gcd(a, b)=1}}^{\infty} \frac{1}{\left ( ab \right )^s}\]

Moreover for s>1 we have that

    \begin{align*} \zeta^2(s) &= \left ( \sum_{a=1}^{\infty} \frac{1}{a^s} \right )^2 \\ &=\sum_{a, b=1}^{\infty} \frac{1}{(ab)^s} \\ &=\sum_{d=1}^{\infty} \sum_{\substack{a, b=1 \\\gcd(a, b)=d}}^{\infty} \frac{1}{\left ( ab \right )^s} \\ &= \sum_{d=1}^{\infty} \frac{1}{d^{2s}} \sum_{\substack{a, b=1 \\\gcd(a, b)=1}}^{\infty} \frac{1}{\left ( ab \right )^s} \\ &= \zeta(2s) F(s) \end{align*}

Hence for s=2 we have that

    \[F(2) = \frac{5}{2}\]

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On the dubious function

The dubious function \Delta:\mathbb{N} \rightarrow \mathbb{N} is defined as follows : \Delta(1)=1 and

    \[\mathbf{\Delta(n) = \sum_{\substack{\mathbf{d \mid n } \\ \mathbf{d \neq n} }} \Delta (d)} \quad \text{for} \;\; n>1\]

Evaluate the sum

    \[\mathcal{S} = \sum_{n=0}^{\infty} \frac{\Delta\left ( 15^n \right )}{15^n}\]

Double summation

Evaluate

    \[\mathcal{S} = \sum_{n=1}^{\infty}\left(n\sum_{k=n}^{\infty}\frac{1}{k^2}-1-\frac{1}{2n}\right)\]

Solution

Coming soon!

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