Double series

Evaluate the series:

\displaystyle \sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k}\sum_{n=0}^{\infty}\frac{1}{k2^n+1}

(Putnam Competition , 2016)

Solution

Since the double series the series converges absolutely we can interchange the summation. Thus:

\begin{aligned} \sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{k} \sum_{n=0}^{\infty} \frac{1}{k2^n+1} &= \sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{k} \sum_{n=0}^{\infty} \int_{0}^{1} x^{k 2^n } \, {\rm d}x\\ &= \sum_{n=0}^{\infty} \int_{0}^{1}\sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{k} x^{2^n k} \, {\rm d}x\\ &= \sum_{n=0}^{\infty} \int_{0}^{1} \log \left ( 1+x^{2^n} \right ) \, {\rm d}x\\ &= \int_{0}^{1}\sum_{n=0}^{\infty} \log \left ( 1+ x^{2^n} \right ) \, {\rm d}x \\ &= \int_{0}^{1} \log \prod_{n=0}^{\infty} \left ( 1+x^{2^n} \right ) \, {\rm d}x \\ &= - \int_{0}^{1} \log \left ( 1-x \right ) \, {\rm d}x \\ &=1 \end{aligned}

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On an entire function

Let n \in \mathbb{N} and f be an entire function. Prove that for any arbitrary positive numbers a, b it holds that:

\displaystyle \frac{\bigintsss_{0}^{2\pi} e^{-i n t}f \left ( z+a e^{it} \right ) \, {\rm d}t}{\bigintsss_{0}^{2\pi} e^{-i n t} f\left ( z + b e^{it} \right ) \, {\rm d}t} = \left ( \frac{a}{b} \right )^n

Solution

Since our function is entire this means that it is holomorphic and can be represented in the form

\displaystyle f(x)=\sum_{m=0}^{\infty} a_m \left ( x-z \right )^m

This series converges uniformly on [0, 2\pi] thus we can interchange summation and integral. Hence:

\begin{aligned} \int_{0}^{2\pi} e^{-in t} f\left ( z + ae^{it} \right ) \, {\rm d}t &= \int_{0}^{2\pi} \sum_{m=0}^{\infty} a_m a^m e^{it \left ( m-n \right )} \, {\rm d}t \\ &= \sum_{m=0}^{\infty} a_m a^m \int_{0}^{2\pi}e^{it \left ( m-n \right )} \, {\rm d}t \\ &= 2 \pi \sum_{m=0}^{\infty} a_m a^m \delta_{mn}\\ &=2 \pi a_n a^n \end{aligned}

where \delta_{mn} is Kronecker’s delta. Similarly for the denominator. Dividing we get the result.

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A trigonometric identity

Prove that

\arctan 1 + \arctan 2 + \arctan 3 = \pi

Solution

It suffices to evaluate the value of

\displaystyle \arctan 2 + \arctan 3 = \frac{3 \pi}{4}

Indeed:

    \begin{align*} \arctan 2 + \arctan 3 &=\arctan \left ( \frac{2+3}{1- 2\cdot 3} \right ) \\ &= \arctan \left ( -1 \right )\\ &= n\pi - \frac{\pi}{4} \end{align*}

The principal value of the quantity we are seeking lies in the interval  (0, \pi). Thus n=1 and consequently we get the result.

Thus:

    \begin{align*} \arctan 1 + \arctan 2 + \arctan 3 &= \frac{\pi}{4} + \frac{3\pi}{4} \\ &= \pi \end{align*}

Note: In general , for positive a,b,c it holds that:

\arctan a + \arctan b + \arctan c = \pi \iff a + b+ c = abc

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A rational number

Let n \in \mathbb{N}. Prove that the number

\mathfrak{n}=\sqrt{\underbrace{1111\cdots11}_{2n} - \underbrace{2222\cdots22}_{n}}

is rational.

Solution

We have successively that

    \begin{align*} \mathfrak{n} &= \sqrt{\underbrace{111\cdots11}_{2n} - \underbrace{222\cdots22}_{n}} \\ &= \sqrt{\sum_{k=0}^{2n-1} 10^{k} -2\sum_{k=0}^{n-1} 10^k}\\ &=\sqrt{\frac{1}{9} \left ( 10^{2n}-1 \right ) - \frac{2}{9} \left ( 10^n -1 \right ) } \\ &= \sqrt{\frac{\left ( 10^n-1 \right )^2}{9}} \\ &=\frac{10^n-1}{3} \\ &=3 \left ( 1 + 10 + 100 + \cdots +10^{n-1} \right ) \\ &= \underbrace{333\cdots33}_{n} \end{align*}

since every number n \in \mathbb{N} has an expansion of the form

\displaystyle \sum_{k=0}^{n-1}a_k 10^k , \; 0\leq a_k \leq 9

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An inequality

Let a, b, c be positive real numbers such that

a+b+c=1

Prove that

\displaystyle \prod_{\text{cyclic}}\left ( \frac{1}{a} + \frac{1}{bc} \right ) > 1728

(Vojtech Jarnik / Second Category / 2016)

Solution

Using the AM – GM inequality we have that

    \begin{align*} \frac{1}{a} + \frac{1}{bc}&= \frac{1}{a} + \frac{1}{3bc} + \frac{1}{3bc} + \frac{1}{3bc} \\ &> \frac{4}{\sqrt[4]{27 ab^3 c^3}} \end{align*}

as well as \displaystyle \left ( \frac{a+b+c}{3} \right )^3 = \frac{1}{27} >abc. Thus:

    \begin{align*} \prod_{\text{cyclic}} \left ( \frac{1}{a} + \frac{1}{bc} \right ) &> 64 \prod_{\text{cyclic}} \frac{1}{\sqrt[4]{27ab^3c^3}} \\ &=\frac{64}{\sqrt[4]{3^9 (abc)^7}} \\ &> \frac{64}{\sqrt[4]{3^9 \left ( 3^{-3} \right )^7}}\\ &= 64 \cdot \sqrt[4]{3^{12}} \\ &= 1728 \end{align*}

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