Hence the original sum equals $\pi^2/16$.

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In summary, the significance of the sum being equal to $\pi^2/16$ lies in its solution to the famous Basel problem and its role in the study of number theory and analysis. The proof for this result involves the use of complex analysis and the concept of a Fourier series. The sum is written as $\pi^2/16$ instead of a decimal approximation for its precision and connection to other mathematical concepts. Other interesting results related to this sum include its representation as an infinite product involving prime numbers and its connection to the zeta function and the area of a quarter circle. While it may not have direct real-world applications, this result has been used to solve other mathematical problems and has connections to various fields of study.
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Ackbach
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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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  • #2
No one solved this week's POTW, which was B-6 from the 2015 Putnam archive. The solution, from artofproblemsolving.com, follows:

[sp] $\newcommand{\ee}{\ell}$
(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*}
[/sp]
 

FAQ: Hence the original sum equals $\pi^2/16$.

What is the significance of the sum being equal to $\pi^2/16$?

The sum being equal to $\pi^2/16$ is significant because it is a famous result in mathematics known as the Basel problem. This sum was first solved by Leonhard Euler in the 18th century and plays a crucial role in the study of number theory and analysis.

How does one prove that the sum is equal to $\pi^2/16$?

The proof for this result involves using complex analysis and the concept of a Fourier series. By representing the sine function as a combination of complex exponentials, one can manipulate the series to show that it converges to $\pi^2/16$.

Why is the sum written as $\pi^2/16$ instead of a decimal approximation?

Writing the sum as $\pi^2/16$ instead of a decimal approximation, such as 0.6154, allows for a more precise and exact representation of the result. The use of $\pi$ also connects this result to other important mathematical concepts and formulas.

What other interesting results are related to the sum equaling $\pi^2/16$?

There are many other interesting results related to this sum, including the fact that it can also be expressed as an infinite product involving prime numbers. Additionally, the value of the sum is connected to the area of a quarter circle and the zeta function evaluated at 2.

How is this result applied in real-world situations?

While this result may not have direct applications in the real world, it has significant implications in the field of mathematics and its applications. It has been used to solve other mathematical problems and has connections to various areas of study, such as physics and engineering.

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