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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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Arithmotheoretic sum

Evaluate the limit:

    \[\ell= \lim_{n \rightarrow +\infty} \frac{1}{n^2} \sum_{m=1}^{n} n \pmod m\]

Solution

    \begin{align*} \ell &=\lim_{n \rightarrow +\infty} \frac{1}{n^2} \sum_{m=1}^{n} n \pmod m \\ &= \lim_{n \rightarrow +\infty} \frac{1}{n^2} \sum_{m=1}^{n} \left ( n - m \left \lfloor \frac{n}{m} \right \rfloor \right ) \\ &=\lim_{n \rightarrow +\infty} \frac{1}{n^2} \left ( n^2 - \sum_{m=1}^{n} m \left \lfloor \frac{n}{m} \right \rfloor \right ) \\ &=1 - \lim_{n \rightarrow +\infty} \frac{1}{n} \sum_{m=1}^{n} \frac{m}{n} \left \lfloor \frac{n}{m} \right \rfloor \\ &= 1 - \int_{0}^{1} x \left \lfloor \frac{1}{x} \right \rfloor \, \mathrm{d}x \end{align*}

However,

    \begin{align*} \int_{0}^{1} x \left \lfloor \frac{1}{x} \right \rfloor \, \mathrm{d}x &\overset{u=1/x}{=\! =\! =\! =\! =\!} \int_{1}^{\infty} \frac{\left \lfloor u \right \rfloor}{u^3} \, \mathrm{d}x \\ &=\sum_{n=1}^{\infty} \int_{n}^{n+1} \frac{n}{x^3} \, \mathrm{d}x \\ &= \frac{1}{2}\sum_{n=1}^{\infty} n \left ( \frac{1}{n^2} - \frac{1}{(n+1)^2} \right )\\ &= \frac{1}{2} \sum_{n=1}^{\infty} \left ( \frac{1}{n} - \frac{n}{(n+1)^2} \right )\\ &= \frac{1}{2} \sum_{n=1}^{\infty} \left ( \frac{1}{n} - \frac{1}{n+1} + \frac{1}{(n+1)^2} \right )\\ &=\frac{1}{2}\cancelto{1}{\sum_{n=1}^{\infty} \left ( \frac{1}{n} - \frac{1}{n+1} \right )}+ \frac{1}{2} \sum_{n=1}^{\infty} \frac{1}{(n+1)^2} \\ &=\frac{1}{2} + \frac{1}{2}\sum_{n=2}^{\infty} \frac{1}{n^2} \\ &=\frac{1}{2} + \frac{1}{2} \left ( \frac{\pi^2}{6} - 1 \right ) \\ &=\frac{\pi^2}{12} \end{align*}

Hence

    \[\ell = 1 - \frac{\pi^2}{12}\]

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Double binomial sum

Evaluate the double sum

    \[\mathcal{S} = \sum_{i=0}^m \sum_{j=0}^n \frac{\binom{m}{i}\binom{n}{j}}{\binom{m+n}{i+j}}\]

Solution

We sum diagonally , hence:

    \begin{align*} \sum_{i=0}^{m}\sum_{j=0}^{n}\frac{\binom{m}{i}\binom{n}{j}}{\binom{m+n}{i+j}}&\overset{i+j=N}{=\! =\! =\! =\!}\sum_{N=0}^{m+n}\sum_{i=0}^m\frac{\binom{m}{i}\binom{n}{N-i}}{\binom{m+n}{N}}\\ &\stackrel{(1)}{=}\sum_{N=0}^{m+n}\sum_{i=0}^{N}\frac{\binom{m}{i}\binom{n}{N-i}}{\binom{m+n}{N}}\\ &=\sum_{N=0}^{m+n}\binom{m+n}{N}^{-1}\sum_{i=0}^{N}\binom{m}{i}\binom{n}{N-i}\\ &\stackrel{(2)}{=} \sum_{N=0}^{m+n}\binom{m+n}{N}^{-1}\binom{m+n}{N}\\ &=m+n+1 \end{align*}

(1): For i=m+1,\ldots,N it holds \binom{m}{i}=0.

(2): \displaystyle \sum_{i=0}^{N}\binom{m}{i}\binom{n}{N-i}=[x^N](1+x)^{m+n}=\binom{m+n}{N}.

Conjecture: Does the following equality

    \[\sum_{a_1=0}^{b_1}\sum_{a_2=0}^{b_2}\cdots\sum_{a_n=0}^{b_n}\frac{\binom{b_1}{a_1}\binom{b_2}{a_2}\cdots\binom{b_n}{a_n}}{\binom{b_1+b_2+\cdots+b_n}{a_1+a_2+\cdots+a_n}}=b_1+b_2+\cdots+b_n+1\]

hold?

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A double factorial sum

Prove that

    \[\sum_{n=0}^{\infty} \frac{1}{(2n)!!}=\sqrt{e}\]

Solution

We note that

    \begin{align*} \sum_{n=0}^{\infty} \frac{1}{\left ( 2n \right )!!} &= \sum_{n=0}^{\infty} \frac{1}{\left ( 2n \right )\left ( 2n-2 \right )\left ( 2n-4 \right )\cdots 4 \cdot 2} \\ &=\sum_{n=0}^{\infty} \frac{1}{2^n n\left (n- 1 \right ) \left ( n-2 \right ) \cdots 1} \\ &=\sum_{n=0}^{\infty} \frac{1}{2^n n!} \\ &= \sqrt{e} \end{align*}

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