An integral

Let n \in \mathbb{Z}_{\geq 0}. Evaluate the integral

    \[\mathcal{J}_n = \int_0^\pi \frac{1- \cos nx}{1-\cos x}\, \mathrm{d}x\]

Solution

We note that

    \begin{align*} \frac{\mathcal{J}_{n+1}+\mathcal{J}_{n-1}}{2} &=\int_{0}^{\pi}\frac{2-\cos (n+1)x-\cos (n-1)x}{2(1-\cos x)}\, \mathrm{d}x \\ &= \int_{0}^{\pi}\frac{1-\cos n x \cos x}{1-\cos x} \, \mathrm{d}x \\ &=\int_{0}^{\pi}\frac{\left ( 1-\cos n x \right )+\cos n x(1-\cos x)}{1-\cos x}\, \mathrm{d}x \\ &=\mathcal{J}_n+\int_{0}^{\pi} \cos n x\, \mathrm{d}x \\ &=\mathcal{J}_n \end{align*}

This shows that we have a geometric progression. Since \mathcal{J}_0 = 0 , \mathcal{J}_1 = \pi it follows that \mathcal{J}_n = n \pi.

Read more

On a generating function

Let \mathcal{H}_n denote the n-th harmonic number and \mathcal{H}_n^{(2)} the n-th harmonic number of order 2 , namely \mathcal{H}_n^{(2)} = \sum \limits_{k=1}^{n} \frac{1}{k^2}. Prove that

    \[\sum_{n=1}^\infty \left( H_n^2-H_n^{(2)} \right)x^{n}=\frac{\ln^2(1-x)}{1-x}\]

Solution

Lemma: Let \{a_n\}_{n \in \mathbb{N}} be a sequence such tht a_0=0. Then

    \[\sum_{n=1}^\infty a_n x^n=\frac{1}{1-x} \sum_{n=1}^\infty (a_n-a_{n-1}) x^n\]

Proof: We have successively:

    \begin{align*} \sum_{n=0}^\infty a_n x^n&=\left( \frac{1}{1-x}-\frac{x}{1-x} \right) \sum_{n=0}^\infty a_n x^n\\ &=\frac{1}{1-x}\sum_{n=0}^\infty a_n x^n-\frac{1}{1-x} \sum_{n=0}^\infty a_n x^{n+1}\\ &=\frac{1}{1-x}\sum_{n=0}^\infty a_n x^n-\frac{1}{1-x}\sum_{n=1}^\infty a_{n-1}x^{n} \end{align*}

Thus,

    \[\sum_{n=1}^\infty a_n x^n=\frac{1}{1-x} \sum_{n=1}^\infty (a_n-a_{n-1}) x^n\]

Hence for the original problem if we let a_n = \mathcal{H}_n^2 then

    \begin{align*} \sum_{n=1}^{\infty} \mathcal{H}_n^2 x^n &= \frac{1}{1-x}\sum_{n=1}^{\infty} \left(\mathcal{H}_n^2-\mathcal{H}_{n-1}^2 \right) x^n\\ &=\frac{1}{1-x} \sum_{n=1}^{\infty} \left( \frac{2\mathcal{H}_n}{n}-\frac{1}{n^2} \right) x^n\\ &=\frac{2}{1-x} \sum_{n=1}^{\infty} \frac{\mathcal{H}_n}{n} x^n-\frac{\mathrm{Li}_2(x)}{1-x}\\ &=\frac{2}{1-x}\left(\mathrm{Li}_2(x) + \frac{\ln^2(1-x)}{2} \right)-\frac{\mathrm{Li}_2(x)}{1-x}\\ &=\frac{\ln^2(1-x)}{1-x}+\frac{\mathrm{Li}_2(x)}{1-x}\\ &=\frac{\ln^2(1-x)}{1-x}+\sum_{n=1}^{\infty} \mathcal{H}_n^{(2)}x^n \end{align*}

Read more

Harmonic series

Let \mathcal{H}_n denote the n-th harmonic number. Prove that

    \[\sum_{n=1}^{\infty}\frac{\mathcal{H}_n}{n(n+1)(n+2)} = \frac{\pi^2}{12} - \frac{1}{2}\]

Solution

We recall the series \displaystyle \sum_{n=1}^{\infty} \mathcal{H}_n x^n = - \frac{\ln (1-x)}{1-x}. Integrating we get

    \[\sum_{n=1}^\infty \frac{\mathcal{H}_n}{n+1}x^{n+1} = \frac{1}{2}\ln^2(1-x)\]

The last takes the form

    \[\sum_{n=1}^\infty \frac{\mathcal{H}_n}{n}x^{n} = \frac{1}{2}\,\ln^2(1-x) + \mathrm{Li}_2(x)\]

where \mathrm{Li}_2 denotes the dilogarithm function. Thus,

    \begin{align*} \sum_{n=1}^{\infty} \frac{\mathcal{H}_n}{n \left ( n+1 \right ) \left ( n+2 \right )} &= \sum_{n=1}^{\infty} \frac{\mathcal{H}_n}{n \left ( n+1 \right )} \int_{0}^{1} x^{n+1} \, \mathrm{d}x \\ &= \int_{0}^{1} \sum_{n=1}^{\infty} \frac{\mathcal{H}_n}{n \left ( n+1 \right )} x^{n+1} \,\mathrm{d}x \\ &= \int_{0}^{1} \sum_{n=1}^{\infty} \mathcal{H}_n \left ( \frac{1}{n} - \frac{1}{n+1} \right ) x^{n+1} \, \mathrm{d}x \\ &= \frac{1}{2}\int_{0}^{1} x \ln^2 (1-x)\, \mathrm{d}x +\int_0^1 x \mathrm{Li}_2(x) \, \mathrm{d}x \\ &\quad \quad \quad \quad -\frac{1}{2} \int_0^1 \ln^2(1-x) \, \mathrm{d}x \\ &=\frac{1}{2} \cdot \frac{7}{4} + \frac{\pi^2}{12} - \frac{9}{24} - 2 \cdot \frac{1}{2} \\ &= \frac{\pi^2}{12} - \frac{1}{2} \end{align*}

Read more

Pell-Lucas series

The Pell-Lucas numbers \mathcal{Q}_n satisfy \mathcal{Q}_0 = \mathcal{Q}_1 =2 and \mathcal{Q}_n = 2\mathcal{Q}_{n-1} +\mathcal{Q}_{n-2} for n \geq 2. Prove that

    \[\sum_{n=1}^{\infty} \arctan \frac{2}{\mathcal{Q}_n} \arctan \frac{2}{\mathcal{Q}_{n+1}} = \frac{\pi^2}{32}\]

\varphi summation

Let \varphi denote the Euler’s function. Prove that

    \[\sum_{k=1}^{\infty} \varphi(k) \left \lfloor \frac{n}{k} \right \rfloor = \frac{n \left ( n+1 \right )}{2}\]

Solution

The key idea is to rewrite the floor as a sum involving divisors. Hence,

    \begin{align*} \sum_{k=1}^{\infty} \varphi(k) \left \lfloor \frac{n}{k} \right \rfloor &= \sum_{k=1}^{\infty} \varphi(k) \sum_{\substack{m \leq n \\ k \mid n}} 1 \\ &= \sum_{k=1}^{\infty} \sum_{\substack{m \leq n \\ k \mid n}} \varphi(k) \\ &= \sum_{m=1}^{n} \sum_{k \mid m} \varphi(k) \\ &= \sum_{m=1}^{n} m \\ &= \frac{n \left ( n+1 \right )}{2} \end{align*}

Read more

Donate to Tolaso Network