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On a polygamma product series

Let $\psi^{(1)}$ denote the trigamma function. Prove that

$\sum_{n=1}^{\infty} \frac{\psi^{(1)}(n) \psi^{(1)} (n+1)}{n^2} = \frac{\pi^6}{840}$

(Seraphim Tsipelis)

Solution

We are quoting several lemmata in order to prove this result.

Lemma 1: It holds that $\displaystyle \psi^{(1)} (n+1) = \psi^{(1)} (n) - \frac{1}{n^2}$ .

Proof:

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

Lemma 2: It holds that $\displaystyle \zeta(2) - \sum_{k=1}^{n} \frac{1}{k^2} = \psi^{(1)} (n+1)$ .

Proof:

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

Lemma 3: We define $\displaystyle \mathcal{H}_n^{(s)} = \sum_{k=1}^{n} \frac{1}{k^s}$ with $\mathbb{N} \ni s \geq 1$ .

Lemma 4: It holds that $\displaystyle \sum_{n=1}^{\infty} \frac{\mathcal{H}_n^{(k)}}{n^k} = \frac{\zeta(2k)+ \zeta^2(k)}{2}$ .

Proof:

$\begin{align*} \zeta^2(k) &= \sum_{n=1}^\infty \sum_{m=1}^\infty \frac{1}{n^km^k} \\&= 2\sum_{1\leq m\leq n} \frac{1}{(nm)^k} - \sum_{m=1}^\infty\frac{1}{(n^k)^2}\\ &=2\sum_{n=1}^\infty \frac{1}{n^k}\sum_{m=1}^n \frac{1}{m^k} - \zeta(2k)\\ &=2\sum_{n=1}^\infty \frac{H_n^{(k)}}{n^k} - \zeta(2k) \end{align*}$

and the result follows.

Lemma 5: It holds that $\displaystyle \sum_{n=1}^{\infty} \frac{\mathcal{H}_n^{(2)}}{n^4} =\zeta^2(3) - \frac{\zeta(6)}{3}$ .

Lemma 6: It holds that $\displaystyle \sum_{n=1}^{\infty} \frac{\left (\mathcal{H}_n^{(2)} \right )^2}{n^2} = \zeta^2(3) + \frac{19}{24} \zeta(6)$ .

Proof:

The proof relies on Multiple Zeta functions. Since it holds that

$\left(\mathcal{H}_n^{(2)}\right)^2=\mathcal{H}_n^{(4)}+2\sum_{k=1}^{n}\frac{\mathcal{H}_{k-1}^{(2)}}{k^2}$

then we have that

$\begin{aligned} \sum_{n=1}^{\infty} \frac{\left( \mathcal{H}_n^{(2)} \right)^2}{n^2} &=\sum_{n=1}^{\infty}\frac{\mathcal{H}_n^{(4)}}{n^2}+ 2\sum_{n=1}^{\infty}\frac{1}{n^2}\sum_{k=1}^{n}\frac{\mathcal{H}_{k-1}^{(2)}}{k^2}\\ &=\left(\zeta(6)+\zeta(2,4)\right)+2(\zeta(2,2,2)+\zeta(4,2))\\ &=\zeta(6)+\left(\frac{25}{12}\zeta(6)-\zeta^2 (3)\right)+2\left(\frac{\zeta^3(2)}{6}+\frac{\zeta(6)}{3}-\frac{\zeta(2)\zeta(4)}{2}\right)\\ &\qquad+2\left(\zeta^2(3)-\frac{4\zeta(6)}{3}\right)\\ &=\zeta^2(3)+\frac{19\pi^6}{22680}\\ &=\zeta^2(3) + \frac{19 \zeta(6)}{24} \end{aligned}$

Now returning to the question in hand we have that

$\begin{aligned} \sum_{n=1}^{\infty} \frac{\psi^{(1)}(n) \psi^{(1)}(n+1)}{n^2} &= \sum_{n=1}^{\infty} \frac{\left (\psi^{(1)}(n+1) + \frac{1}{n^2} \right ) \psi^{(1)}(n+1)}{n^2} \\ &= \sum_{n=1}^{\infty} \frac{\left ( \psi^{(1)}(n+1) \right )^2}{n^2}+ \sum_{n=1}^{\infty} \frac{\psi^{(1)}(n+1) }{n^4}\\ &= \underbrace{\sum_{n=1}^{\infty} \frac{\left ( \zeta(2) - \mathcal{H}_n^{(2)} \right )^2}{n^2}}_{\mathcal{S}_1} + \underbrace{\sum_{n=1}^{\infty} \frac{\zeta(2) - \mathcal{H}_n^{(2)}}{n^4}}_{\mathcal{S}_2}\ \end{aligned}$

and what remains in order to complete the exercise is the evaluation of those two last Euler sums. We proceed with the second one.

$\begin{aligned} \mathcal{S}_2 &= \sum_{n=1}^{\infty} \frac{\zeta(2) - \mathcal{H}_n^{(2)}}{n^4} \\ &= \sum_{n=1}^{\infty} \frac{\zeta(2)}{n^4} - \sum_{n=1}^{\infty} \frac{\mathcal{H}_n^{(2)}}{n^4} \\ &= \zeta(2) \zeta(4) - \zeta^2(3) + \frac{\zeta(6)}{3} \\ &= \frac{\pi^6}{540} - \zeta^2(3) + \frac{\pi^6}{2835} \end{aligned}$

while for the first one

$\begin{align*} \mathcal{S}_1 &= \sum_{n=1}^{\infty} \frac{\left ( \zeta(2) - \mathcal{H}_n^{(2)} \right )^2}{n^2} \\ &= \sum_{n=1}^{\infty} \frac{\zeta^2(2) - 2 \zeta(2) \mathcal{H}_n^{(2)} + \left (\mathcal{H}_n^{(2)} \right )^2}{n^2}\\ &=\zeta^3(2) - 2 \zeta(2) \sum_{n=1}^{\infty} \frac{\mathcal{H}_n^{(2)}}{n^2} + \sum_{n=1}^{\infty} \frac{\left (\mathcal{H}_n^{(2)} \right )^2}{n^2} \\ &= \zeta^3(2) - 2 \zeta(2) \cdot \frac{7 \zeta(4)}{4} + \sum_{n=1}^{\infty} \frac{\left (\mathcal{H}_n^{(2)} \right )^2}{n^2} \\ &= \frac{\pi^6}{216} - \frac{7 \zeta(2) \zeta(4)}{2} + \sum_{n=1}^{\infty} \frac{\left (\mathcal{H}_n^{(2)} \right )^2}{n^2} \\ &= \frac{\pi^6}{216} - \frac{7 \pi^6}{1080} + \sum_{n=1}^{\infty} \frac{\left (\mathcal{H}_n^{(2)} \right )^2}{n^2} \\ &= - \frac{\pi^6}{540} + \sum_{n=1}^{\infty} \frac{\left (\mathcal{H}_n^{(2)} \right )^2}{n^2} \\ &= - \frac{\pi^6}{540} + \zeta^2(3) + \frac{19 \zeta(6)}{24} \end{align*}$

Combining the results we get what we want.

A series with least common multiple

Let $\{X_n\}_{n \in \mathbb{N}}$ be a strictly increasing sequence of positive numbers. For all $n \geq 1$ denote as $W_n$ the least common multiple of the first $n$ terms $X_1, X_2, \dots, X_n$ of the sequence. Prove that , as $n \rightarrow +\infty$ , the following sum converges

$\mathcal{S} = \frac{1}{W_1} + \frac{1}{W_2} + \cdots + \frac{1}{W_n}$

Solution

This is a result due to Paul Erdös stating that if $X_1, X_2, \dots, X_n$ are natural numbers such that $1 \leq X_1 < X_2 < \cdots < X_n$ then

$\displaystyle \frac{1}{\lcm(X_0,X_1)}+\frac{1}{\lcm(X_1,X_2)}+\cdots+\frac{1}{\lcm(X_{n-1},X_n)}\leq 1-\frac{1}{2^n}$

and the original question follows since the sum we seek is less or equal to $\displaystyle 1 - \frac{1}{2^n}$ .

However, we are presenting another proof. Denote as $W(n)$ the average order of the numbers $W_n$ , i.e.,

$W (n) = \frac {1} {n} \sum_{k = 1}^{n} W_k.$

For any $k$ we have $W_{k + 1} = W_k \cdot m_k$ where $m_k$ is the product of primes not present in the factorization of $X_1, X_2, \cdots, X_k$ . Note that $m_k$ are squarefree integers. Note also that it may be an empty product, i.e., $m_k = 1$ . Then

$\sum_{k = 1}^{n} W_k = W_1 \cdot \sum_{k = 0}^{n - 1} \prod_{j = 0}^{k} m_j.$

It is easy to see (and show by induction) that $\prod \limits_{j = 0}^{k} m_j > 2^k$ so we have

$\sum_{k = 1}^{n} W_k > W_1 \cdot \sum_{k = 0}^{n - 1} 2^k = (2^n - 1) W_1.$

Hence, $W (n) > \frac {2^n - 1} {n} W_1.$ Consequently, we have

$\sum_{n = 1}^{\infty} \frac {1} {W (n)} < \frac {1} {W_1} \sum_{n = 1}^{\infty} \frac {n} {2^n - 1}$

So the sum of reciprocals of $W (n)$ converges. Then, by Cesàro summation, we see that

$\sum_{n = 1}^{\infty} \frac {1} {W_n}$

also converges.

Divisibility

Prove that the product of $n$ consecutive positive integers divides $n!$ .

Solution

The number of different combinations of $N+n$ over $N$ is of course an integer and equals to

$\begin{pmatrix} N+n\\ N \end{pmatrix} = \frac{(N+1)(N+2)...(N+n)}{n!}$

and the result follows.

Any questions?

Inequality in acute triangle

Let $ABC$ be an acute triangle. Prove the following inequality

$\sum \cot^2 {\rm A} \cot^2 {\rm B} \geq \frac{\sum \cos^2 {\rm A}}{\sum \sin^2 {\rm A}}$

Solution

The solution can be found at cut -the – knot.

Inequality in a triangle

Given a triangle $ABC$ let $m_a , m_b , m_c$ denote the median points of the sides $a , b, c$ respectively. Prove that

$\left ( m_a + m_b + m_c \right )^2 + 4 r \left ( R - 2r \right ) \leq \left ( 4 R + r \right )^2$

where $R$ denotes the the circumradius and $r$ the inradius respectively.

(Adil Abdullayev / RMM)

Solution [Soumava Chakraborty]

We have successively

$\begin{align*} \left ( \sum m_a \right )^2 &=\sum m_a^2 + 2 \sum m_a m_b \\ &=\frac{3}{4} \sum a^2 + 2 \sum m_a m_b \\ &\!\!\!\!\!\!\!\!\!\!\!\!\!\overset{m_am_b \leq \frac{2c^2+ab}{4}}{\leq } \frac{3}{4} \sum a^2 + \frac{1}{2} \sum \left ( 2c^2 + ab \right )\\ &= \frac{3}{4} \sum a^2 + \sum a^2 + \frac{1}{2} \sum ab\\ &= \frac{7}{4} \sum a^2 + \frac{1}{2} \sum ab \\ &= \frac{1}{4} \left [ \sum a^2 + 2 \sum ab + 6 \sum a^2 \right ]\\ &= \frac{4s^2 + 12 \left ( s^2-4Rr -r^2 \right )}{4} \\ &=4s^2 -12 Rr -3r^2 \\ &\!\!\!\!\!\!\!\!\!\overset{\text{Gerretsen}}{\leq } 4 \left ( 4R^2 + 4Rr + 3r^2 \right ) -12 Rr -3r^2 \\ &=16R^2 + 8Rr + r^2 -4Rr + 8r^2 \\ &=\left ( 4R+r \right )^2 - 4r \left ( R - 2r \right ) \end{align*}$

In all the above $s$ denotes the semiperimeter of the triangle. More on Gerretsen’s inequality can be found at this link .