Problem Of The Week # 287 - Oct 31, 2017

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Ackbach

Indicium Physicus
Staff member
Here is this week's POTW:

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For each positive integer $k$, let $A(k)$ be the number of odd divisors of $k$ in the interval $[1, \sqrt{2k}\,)$. Evaluate
$\sum_{k=1}^\infty (-1)^{k-1} \frac{A(k)}{k}.$

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Ackbach

Indicium Physicus
Staff member
No one solved this week's POTW, which was B-6 from the 2015 Putnam archive. The solution, from artofproblemsolving.com, follows:

$\newcommand{\ee}{\ell}$
(from \url{artofproblemsolving.com})
We will prove that the sum converges to $\pi^2/16$. Note first that the sum does not converge absolutely, so we are not free to rearrange it arbitrarily. For that matter, the standard alternating sum test does not apply because the absolute values of the terms does not decrease to 0, so even the convergence of the sum must be established by hand.

Setting these issues aside momentarily, note that the elements of the set counted by $A(k)$ are those odd positive integers $d$ for which $m = k/d$ is also an integer and $d < \sqrt{2dm}$; if we write $d = 2\ee-1$, then the condition on $m$ reduces to $m \geq \ee$. In other words, the original sum equals
$S_1 := \sum_{k=1}^\infty \sum_{{\ee \geq 1, m \geq \ee}\atop{k = m(2\ee-1)}} \frac{(-1)^{m-1}}{m(2\ee-1)},$
and we would like to rearrange this to
$S_2 := \sum_{\ee=1}^\infty \frac{1}{2\ee-1} \sum_{m=\ee}^\infty \frac{(-1)^{m-1}}{m},$
in which both sums converge by the alternating sum test. In fact a bit more is true: we have
$\left| \sum_{m=\ee}^\infty \frac{(-1)^{m-1}}{m} \right| < \frac{1}{\ee},$
so the outer sum converges absolutely. In particular, $S_2$ is the limit of the truncated sums
$S_{2,n} = \sum_{\ee(2\ee-1) \leq n} \frac{1}{2\ee-1} \sum_{m=\ee}^\infty \frac{(-1)^{m-1}}{m}.$
To see that $S_1$ converges to the same value as $S_2$, write
$S_{2,n} - \sum_{k=1}^n (-1)^{k-1} \frac{A(k)}{k} = \sum_{\ee(2\ee-1) \leq n} \frac{1}{2\ee-1} \sum_{m=\lfloor \frac{n}{2\ee-1}+1 \rfloor}^\infty \frac{(-1)^{m-1}}{m}.$
The expression on the right is bounded above in absolute value by the sum $\sum_{\ee(2\ee-1) \leq n} \frac{1}{n}$, in which the number of summands is at most $\sqrt{n}$ (since $\sqrt{n}(2\sqrt{n}-1)\geq n$), and so the total is bounded above by $1/\sqrt{n}$. Hence the difference converges to zero as $n \to \infty$; that is, $S_1$ converges and equals $S_2$.

We may thus focus hereafter on computing $S_2$. We begin by writing
$S_2 = \sum_{\ee=1}^\infty \frac{1}{2\ee-1} \sum_{m=\ee}^\infty (-1)^{m-1} \int_0^1 t^{m-1}\,dt.$
Our next step will be to interchange the inner sum and the integral, but again this requires some justification.

Lemma 1

Let $f_0, f_1, \dots$ be a sequence of continuous functions on $[0,1]$ such that for each $x \in [0,1]$, we have
$f_0(x) \geq f_1(x) \geq \cdots \geq 0.$
Then
$\sum_{n=0}^\infty (-1)^n \int_0^1 f_n(t)\,dt = \int_0^1 \left( \sum_{n=0}^\infty (-1)^n f_n(t) \right)\,dt$
provided that both sums converge.

Proof:

Put $g_n(t) = f_{2n}(t) - f_{2n+1}(t) \geq 0$; we may then rewrite the desired equality as
$\sum_{n=0}^\infty \int_0^1 g_n(t) \,dt = \int_0^1 \left( \sum_{n=0}^\infty g_n(t) \right)\,dt,$
which is a case of the Lebesgue monotone convergence theorem.

By Lemma 1, we have
\begin{align*}
S_2 &= \sum_{\ee=1}^\infty \frac{1}{2\ee-1} \int_0^1 \left( \sum_{m=\ee}^\infty (-1)^{m-1} t^{m-1} \right) \,dt \\
&= \sum_{\ee=1}^\infty \frac{1}{2\ee-1} \int_0^1 \frac{(-t)^{\ee-1}}{1+t} \,dt.
\end{align*}
Since the outer sum is absolutely convergent, we may freely interchange it with the integral:
\begin{align*}
S_2 &= \int_0^1 \left(
\sum_{\ee=1}^\infty \frac{1}{2\ee-1} \frac{(-t)^{\ee-1}}{1+t} \right)\,dt \\
&= \int_0^1 \frac{1}{\sqrt{t}(1+t)} \left( \sum_{\ee=1}^\infty \frac{(-1)^{\ee-1} t^{\ee-1/2}}{2\ee-1} \right) \,dt \\
&= \int_0^1 \frac{1}{\sqrt{t}(1+t)} \arctan(\sqrt{t})\,dt \\
&= \int_0^1 \frac{2}{1+u^2} \arctan(u)\,du \qquad (u = \sqrt{t}) \\
&= \arctan(1)^2 - \arctan(0)^2 = \frac{\pi^2}{16}.
\end{align*}

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