On a sequence and series

Let x_n be a sequence of real numbers , \mathcal{S} = \sum \limits_{n =1}^{\infty} x_n and \ell = \lim \limits_{ n \rightarrow +\infty} n x_n.

  1. Prove that if \mathcal{S} converges and \ell exists ( finite or infinite ) then \ell =0.
  2. Give an example where x_n>0 , \mathcal{S} converges but \ell does not exist.
  3. Give an example of a decreasing sequence x_n , \ell =0 but \mathcal{S} diverges.
  4. Prove that if \mathcal{S} converges and x_n is decreasing then

        \[\sum_{n=1}^{\infty} n \left ( x_n - x_{n+1} \right ) = \mathcal{S}\]

Solution

  1. Let \mathcal{S}_n be the sequence of partial sums , then \mathcal{S}_n \rightarrow \mathcal{S}. It follows from Cesaro that \displaystyle \frac{\mathcal{S}_1 + \mathcal{S}_2 + \cdots + \mathcal{S}_n}{n} \rightarrow \mathcal{S}. Hence

        \[\frac {nx_1+(n-1)x_2+...+2x_{n-1}+x_n}{n} \rightarrow \mathcal{S}\]

    From the assumption we have that

        \[\frac{n+1}{n} \left(x_1+ \cdots +x_n \right) = \frac {n+1}{n} \mathcal{S}_n \rightarrow \mathcal{S}\]

    Substracting these two we have that

        \[\frac {x_1+2 x_2+...+(n-1)x_{n-1}+nx_n}{n} \rightarrow 0\]

    But since n x_n \rightarrow \ell it follows from Cesaro and the uniqueness of the limit that the last sum tends to \ell.

  2. One such example could be \displaystyle x_n = \frac{1}{n^2} if n \neq \frac{1}{2^k} and \displaystyle x_{2^n} =\frac{1}{2^n} otherwise. Now, the series converges but since n \cdot \frac{1}{n^2} \rightarrow 0 and 2^n x_{2^n} \rightarrow 1 the sequence n x_n does not converge.
  3. The classic example is \displaystyle x_n =\frac{1}{n \ln n}.
  4. Since x_n is decreasing it follows that n x_n \rightarrow 0. Let t_n be the partial sum of the LHS. It follows that

        \begin{align*} t_n &=(x_1-x_2)+2(x_2-x_3)+\cdots +n(x_n-x_{n+1}) \\ &= (x_1+x_2+...+x_n)-nx_{n+1}\\ &=s_n-\dfrac {n+1}{n} (n+1)x_{n+1} \\ &\rightarrow \mathcal{S}-0 \end{align*}

    The result now follows.

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Vanishing double summation

Let s>2. Evaluate the series

    \[\mathcal{S} = \sum_{(m,n)\in\mathbb{Z}^2\setminus\{(0,0)\}}\frac{m^2+4mn+n^2}{(m^2+mn+n^2)^s}\]

Solution

Let \omega=e^{2\pi i/3} and z=m-n\omega, then

    \[\sum_{(m,n)\in\mathbb{Z}^2\setminus\{(0,0)\}}\frac{m^2+4mn+n^2}{(m^2+mn+n^2)^s}= -\sum_{\substack{z\in\mathbb Z[\omega]\\ z\ne0}}\frac{\overline{\omega}z^2+\omega\overline{z}^2}{(z\overline{z})^s}=0\]

since the sum over every triple z,\omega z,\overline{\omega}z vanishes (one should also check that the sum absolutely converges but that’s straightforward by abelian summation).

The problem was first proposed on AoPS . There is a second solution though on MSE.

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A product

Let \left| \alpha \right|<1. Prove that:

    \[\prod_{n=0}^{\infty} \left ( 1 + \alpha^{3n} + \alpha^{2\cdot 3^n} \right ) = \frac{1}{1-\alpha}\]

Solution

The product eventually telescopes;

    \begin{align*} \prod _{n = 0}^\infty \left(\alpha^{0 \cdot 3^n} + \alpha^{1 \cdot 3^n} + \alpha^{2 \cdot 3^n} \right) &= \prod^\infty _{n = 0} \frac{1 - (\alpha^{3^n})^3}{1-\alpha^{3^n}} \\ &= \prod ^\infty _{n = 0} \frac{1 - \alpha^{3^{n+1}}}{1 - \alpha^{3^n}} \\ &= \frac{1-\alpha^{3^1}}{1 - \alpha^{3^0}} \cdot \frac{1-\alpha^{3^2}}{1 - \alpha^{3^1}} \cdot \frac{1-\alpha^{3^3}}{1 - \alpha^{3^2}} \cdots \\ &= \frac{1}{1-\alpha} \end{align*}

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Arctan integral

Prove that

    \[\int_0^{\infty}\frac{\arctan x -x e^{-x}}{x^2}\,\mathrm{d}x=1+\gamma\]

where \gamma denotes the Euler – Mascheroni constant.

Solution

Beginning by parts we have,

    \begin{align*} \int_{0}^{\infty} \frac{\arctan x- xe^{-x}}{x^2} \, \mathrm{d}x &= \left [ -\frac{\arctan x - xe^{-x}}{x} \right ]_0^\infty + \\ &\quad \quad \quad + \int_{0}^{\infty} \frac{1}{x} \left ( \frac{1}{x^2+1} +e^{-x} (x-1) \right ) \, \mathrm{d}x \\ &=\int_{0}^{\infty} \left ( \frac{1}{x\left ( x^2+1 \right )} + \frac{e^{-x}(x-1)}{x} \right ) \, \mathrm{d}x \\ &=\int_{0}^{\infty} \left ( \frac{1}{x} - \frac{x}{x^2+1} + e^{-x} - \frac{e^{-x}}{x} \right ) \, \mathrm{d}x \end{align*}

However,

\begin{aligned} \int_{\epsilon}^{M} \left ( \frac{1}{x} - \frac{x}{x^2+1} - \frac{e^{-x}}{x} \right ) \, \mathrm{d}x &= \left ( \ln M - \ln \epsilon - \frac{\ln \left ( M^2+1 \right ) - \ln \left ( \epsilon^2+1 \right ) }{2} \right ) - \\ &\quad \quad \quad - \int_{\epsilon}^{M} \frac{e^{-x}}{x} \, \mathrm{d}x \\ &=\left ( \ln M - \ln \epsilon - \frac{\ln \left ( M^2+1 \right ) - \ln \left ( \epsilon^2+1 \right ) }{2} \right ) - \\ &\quad \quad \quad - \left [ \ln x e^{-x} \right ]_\epsilon^{M}  - \int_{\epsilon}^{M} e^{-x} \ln x \, \mathrm{d}x \\ &=\left ( \ln M - \frac{\ln \left ( M^2+1 \right )}{2} \right ) + \frac{\ln \left ( \epsilon^2+1 \right )}{2} - \\ &\quad \quad \quad \quad -\ln M e^{-M} + \left (\ln \epsilon \; e^{-\epsilon} - \ln \epsilon \right ) - \\ &\quad \quad \quad \quad  -\int_{\epsilon}^{M} e^{-x} \ln x \, \mathrm{d}x \end{aligned}

The result now follows taking \epsilon \rightarrow 0^+ and M \rightarrow +\infty.

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Bessel function integral

Let J_0 denote the Bessel function of the first kind. Prove that

    \[\int_0^\infty J_0(x) \, \mathrm{d}x=1\]

Solution

We recall that

    \[J_0(x)=\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{4^n (n!)^2}\]

Hence,

    \begin{align*} \int_{0}^{\infty}J_0(x)e^{-px}\,\mathrm{d}x &= \sum_{n=0}^{\infty}\frac{(-1)^n}{4^n p^{2n+1}}\binom{2n}{n}\\ &=\frac{1}{p}\sum_{n=0}^{\infty} \binom{-1/2}{n}\frac{1}{p^{2n}} \\ &= \frac{1}{\sqrt{1+p^2}} \end{align*}

Then,

    \[\lim_{p \rightarrow 0^+} \int_{0}^{\infty}J_0(x)e^{-px}\,\mathrm{d}x = \lim_{p \rightarrow 0^+} \frac{1}{\sqrt{1+p^2}} =1\]

Using the fact that the J_0(x) looks like an ‘almost periodic’ function with decreasing amplitude. If we denote by \{\alpha_k\}_{k \geq 0} the zeros of J_0 then \alpha_k \nearrow \infty as k \to \infty and furthermore

    \[\left| \int_{\alpha_k}^{\infty} J_0(x)e^{-px}\,\mathrm{d} x \right| \leq \int_{\alpha_k}^{\alpha_{k+1}} |J_0(x)|e^{-px}\,\mathrm{d} x \rightarrow 0\]

as k \rightarrow \infty for each p \geq 0. So the integral converges uniformly in this case justifying the interchange of limit and integral.

The result follows.

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