A series limit

November 17, 2018 / Leave a comment

Evaluate the limit:

    \[\ell = \lim_{x \rightarrow +\infty}\left(\sum_{n=1}^\infty\left(\frac xn\right)^n\right)^{1/x}\]

Solution

It is quite known that

    \[\left(1+\frac{1}{n} \right)^n \leq e \leq\left(1+\frac{1}{n} \right)^{n+1}\]

Thus,

    \begin{align*} \frac{x}{(n+1)e} &\leq \frac{x}{n+1}\left(\frac{n}{n+1}\right)^n \\ &=\frac{\left(\frac x{n+1}\right)^{n+1}}{\left(\frac xn\right)^n} \\ &=\frac{x}{n}\left(\frac{n}{n+1}\right)^{n+1}\\ &\leq \frac{x}{ne} \end{align*}

Therefore, by induction , for n \geq 1,

    \[\frac{e}{n!}\left(\frac {x}{e} \right)^n \leq \left(\frac {x}{n} \right)^n \leq \frac{x}{(n-1)!}\left(\frac {x}{e} \right)^{n-1}\]

Summing the last equation we get:

    \[e\left(e^{x/e}-1\right) \leq \sum_{n=1}^\infty\left(\frac xn\right)^n \leq xe^{x/e}\]

and raising the last equation to the 1/x power , we get:

    \[\left(e\left(1-e^{-x/e}\right)\right)^{1/x}e^{1/e}\leq \left(\sum_{n=1}^\infty\left(\frac {x}{n} \right)^n\right)^{1/x}\leq x^{1/x}e^{1/e}\]

Thus, by the squeeze theorem the limit is equal to e^{1/e}.

 

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A limit on Dirichlet function

November 13, 2018 / Leave a comment

Let f:\mathbb{R} \rightarrow \mathbb{R} be the Dirichlet function;

    \[f(x) = \left\{\begin{matrix} 1 & , & x\in \mathbb{Q} \\ 0& , & x\in \mathbb{R} \setminus \mathbb{Q} \end{matrix}\right.\]

Evaluate the limit

    \[\ell = \lim_{n \rightarrow +\infty} \frac{1}{n} \left ( f \left ( 1 \right ) + f \left ( \sqrt{2} \right ) + f \left ( \sqrt{3} \right )+ \cdots + f \left ( \sqrt{n} \right ) \right )\]

We simply note that

    \[0 \leq \frac{f \left ( 1 \right ) + f \left ( \sqrt{2} \right ) + f \left ( \sqrt{3} \right )+ \cdots + f \left ( \sqrt{n} \right )}{n} = \frac{\left \lfloor n \right \rfloor}{n} \leq \frac{1}{\sqrt{n}}\]

and the limit follows to be 0. The reason why

    \[f \left ( 1 \right ) + f \left ( \sqrt{2} \right ) + f \left ( \sqrt{3} \right )+ \cdots + f \left ( \sqrt{n} \right ) = \left \lfloor n \right \rfloor\]

is because \sqrt{m} is rational if-f m is a perfect square.

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Series of zeta sum

October 20, 2018 / Leave a comment

Let \zeta denote the Riemann zeta function. Evaluate the series:

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

Solution

Let a \in \mathbb{R} \mid \left| a \right|<2. We are proving the more general result.

    \[\sum_{n=1}^{\infty} a^n \left ( n - \sum_{k=2}^{n} \zeta(k) \right ) = a \left ( \frac{\psi^{(0)}\left ( 2-a \right ) + \gamma}{1-a} + 1 \right )\]

where \psi^{(0)} denotes the digamma function.

First of all, we note that:

    \begin{align*} \sum_{k=2}^{n} \zeta(k) &= \sum_{k=2}^{n} \sum_{m=1}^{\infty} \frac{1}{m^k} \\ &= \sum_{m=1}^{\infty} \sum_{k=2}^{n} \frac{1}{m^k}\\ &=\sum_{m=1}^{\infty} \sum_{k=0}^{n-2} \frac{1}{m^{k+2}} \\ &=\sum_{m=1}^{\infty} \frac{1}{m^2} \sum_{k=0}^{n-2} \frac{1}{m^k} \\ &=\sum_{m=1}^{1} \frac{1}{m^2} \sum_{k=0}^{n-2} \frac{1}{m^k} + \sum_{m=2}^{\infty} \frac{1}{m^2}\sum_{k=0}^{n-2} \frac{1}{m^k} \\ &= n-1 + \sum_{m=2}^{\infty} \frac{1}{m^2} \left ( \frac{1-\frac{1}{m^{n-1}}}{1-\frac{1}{m}} \right ) \\ &= n-1 +\sum_{m=2}^{\infty} \frac{m^{n-1}-1}{\left ( m-1 \right ) m^n} \\ & = n -1 + \sum_{m=1}^{\infty} \frac{1}{m(m+1)} - \sum_{m=2}^{\infty} \frac{1}{(m-1) m^n} \\ &= n - \sum_{m=2}^{\infty} \frac{1}{(m-1) m^n} \end{align*}

Thus,

    \begin{align*} \sum_{n=1}^{\infty} a^n \left ( n - \sum_{k=2}^{n} \zeta(k) \right ) &= \sum_{m=2}^{\infty} \frac{1}{m-1}\sum_{n=2}^{\infty} \left ( \frac{a}{m} \right )^n \\ &=a^2 \sum_{m=2}^{\infty} \frac{1}{(m-1) m \left ( m-a \right )} \\ &=a^2 \sum_{m=2}^{\infty} \frac{1}{m-a} \left ( \frac{1}{m-1} - \frac{1}{m} \right ) \\ &=a^2 \sum_{m=2}^{\infty} \frac{1}{(m-1)(m-a)} -a^2 \sum_{m=2}^{\infty} \frac{1}{m(m-a)} \\ &=a^2 \sum_{m=1}^{\infty} \frac{1}{m(m+1-a)} - a^2 \sum_{m=1}^{\infty} \frac{1}{m(m-a)} + \\ & \quad \quad \quad \quad \quad + \frac{a^2}{1-a} \\ &=\frac{a^2}{1-a} \left ( \psi^{(0)} (2-a) + \gamma \right ) +\\ & \quad \quad \quad \quad \quad + a \left ( \psi^{(0)} (1-a) + \gamma \right )+\frac{a^2}{1-a}\\ &= a \left ( \frac{\psi^{(0)}(2-a) + \gamma}{1-a} + 1 \right ) \end{align*}

due to the reflection formula \displaystyle \psi^{(0)} (z+1) = \psi^{(0)}(z) + \frac{1}{z}.

Side note: If a=1 then the sum equals \zeta(2) - 1.

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Lebesgue measure of Cantor set

October 2, 2018 / Leave a comment

An alternative way to define the Cantor set is the following:

    \[\mathfrak{C}=[0,1] \setminus \bigcup_{n=1}^\infty \;\bigcup_{k=0}^{3^{n-1}-1} \left(\frac{3k+1}{3^n},\frac{3k+2}{3^n}\right)\]

What is the Lebesgue measure of the Cantor set if we consider it as a subset of \mathbb{R}? Is \mathfrak{C} countable?

A divergent series …. or maybe not?

September 23, 2018 / Leave a comment

The number n ranges over all possible powers with both the base and the exponent positive integers greater than n, assuming each such value only once. Prove that:

    \[\sum_{n} \frac{1}{n-1}=1\]

Let us denote by \mathcal{M} the set of positive integers greater than 1 that are not perfect powers ( i.e are not of the form a^p , where a is a positive integer and p \geq 2 ).  Since the terms of the series are positive , we can freely permute them. Thus,

    \begin{align*} \sum_{n} \frac{1}{n-1} &= \sum_{m \in \mathcal{M}} \sum_{k=2}^{\infty} \frac{1}{m^k-1} \\ &= \sum_{m \in \mathcal{M}} \sum_{k=2}^{\infty} \sum_{j=1}^{\infty} \frac{1}{m^{kj}}\\ &=\sum_{m \in \mathcal{M}} \sum_{j=1}^{\infty} \sum_{k=2}^{\infty} \frac{1}{m^{kj}} \\ &= \sum_{m \in \mathcal{M}} \sum_{j=1}^{\infty} \frac{1}{m^j \left ( m^j-1 \right )} \\ &= \sum_{n=2}^{\infty} \frac{1}{n\left ( n-1 \right )} \\ &= \sum_{n=2}^{\infty} \left ( \frac{1}{n-1} - \frac{1}{n} \right )\\ &= 1 \end{align*}

 

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