Proof of $\frac{1}{e^{\pi}+1}+\frac{3}{e^{3\pi}+1}+\frac{5}{e^{5\pi}+1}+\ldots=\frac{1}{24}$

Question

I would like to prove that $\displaystyle\sum_{\substack{n=1\\n\text{ odd}}}^{\infty}\frac{n}{e^{n\pi}+1}=\frac1{24}$.

I found a solution by myself 10 hours after I posted it, here it is:

$$f(x)=\sum_{\substack{n=1\\n\text{ odd}}}^{\infty}\frac{nx^n}{1+x^n},\quad\quad g(x)=\displaystyle\sum_{n=1}^{\infty}\frac{nx^n}{1-x^n},$$

then I must prove that $f(e^{-\pi})=\frac1{24}$. It was not hard to find the relation between $f(x)$ and $g(x)$, namely $f(x)=g(x)-4g(x^2)+4g(x^4)$.

Note that $g(x)$ is a Lambert series, so by expanding the Taylor series for the denominators and reversing the two sums, I get

$$g(x)=\sum_{n=1}^{\infty}\sigma(n)x^n$$

where $\sigma$ is the divisor function $\sigma(n)=\sum_{d\mid n}d$.

I then define for complex $\tau$ the function $$G_2(\tau)=\frac{\pi^2}3\Bigl(1-24\sum_{n=1}^{\infty}\sigma(n)e^{2\pi in\tau}\Bigr)$$ so that $$f(e^{-\pi})=g(e^{-\pi})-4g(e^{-2\pi})+4g(e^{-4\pi})=\frac1{24}+\frac{-G_2(\frac i2)+4G_2(i)-4G_2(2i)}{8\pi^2}.$$

But it is proven in Apostol "Modular forms and Dirichlet Series", page 69-71 that $G_2\bigl(-\frac1{\tau}\bigr)=\tau^2G_2(\tau)-2\pi i\tau$, which gives $\begin{cases}G_2(i)=-G_2(i)+2\pi\\ G_2(\frac i2)=-4G_2(2i)+4\pi\end{cases}\quad$. This is exactly was needed to get the desired result.

Hitoshigoto oshimai !

I find that sum fascinating. $e,\pi$ all together to finally get a rational. This is why mathematics is beautiful!

Thanks to everyone who contributed.

Answer 1

We will use the Mellin transform technique. Recalling the Mellin transform and its inverse

$$ F(s) =\int_0^{\infty} x^{s-1} f(x)dx, \quad\quad f(x)=\frac{1}{2 \pi i} \int_{c-i \infty}^{c+i \infty} x^{-s} F(s)\, ds. $$

Now, let's consider the function

$$ f(x)= \frac{x}{e^{\pi x}+1}. $$

Taking the Mellin transform of $f(x)$, we get

$$ F(s)={\pi }^{-s-1}\Gamma \left( s+1 \right) \left(1- {2}^{-s} \right) \zeta \left( s+1 \right),$$

where $\zeta(s)$ is the zeta function . Representing the function in terms of the inverse Mellin Transform, we have

$$ \frac{x}{e^{\pi x}+1}=\frac{1}{2\pi i}\int_{C}{\pi }^{-s-1}\Gamma \left( s+1 \right) \left( 1-{2}^{-s} \right) \zeta \left( s+1 \right) x^{-s}ds. $$

Substituting $x=2n+1$ and summing yields

$$\sum_{n=0}^{\infty}\frac{2n+1}{e^{\pi (2n+1)}+1}=\frac{1}{2\pi i}\int_{C}{\pi}^{-s-1}\Gamma \left( s+1 \right)\left(1-{2}^{-s} \right) \zeta\left( s+1 \right) \sum_{n=0}^{\infty}(2n+1)^{-s}ds$$

$$ = \frac{1}{2\pi i}\int_{C}{\pi }^{-s-1}\Gamma \left( s+1 \right) \left(1-{2}^{-s} \right)^2\zeta\left( s+1 \right) \zeta(s)ds.$$

Now, the only contribution of the poles comes from the simple pole $s=1$ of $\zeta(s)$ and the residue equals to $\frac{1}{24}$. So, the sum is given by

$$ \sum_{n=0}^{\infty}\frac{2n+1}{e^{\pi (2n+1)}+1}=\frac{1}{24} $$

Notes: 1)

$$ \sum_{n=0}^{\infty}(2n+1)^{-s}= \left(1- {2}^{-s} \right) \zeta \left( s \right). $$

2) The residue of the simple pole $s=1$, which is the pole of the zeta function, can be calculated as

$$ r = \lim_{s=1}(s-1)({\pi }^{-s-1}\Gamma \left( s+1 \right) \left({2}^{-s}-1 \right)^2\zeta\left( s+1 \right) \zeta(s))$$

$$ = \lim_{s\to 1}(s-1)\zeta(s)\lim_{s\to 1} {\pi }^{-s-1}\Gamma \left( s+1 \right) \left({2}^{-s}-1 \right)^2\zeta\left( s+1 \right) = \frac{1}{24}. $$

For calculating the above limit, we used the facts

$$ \lim_{s\to 1}(s-1)\zeta(s)=1, \quad \zeta(2)=\frac{\pi^2}{6}. $$

3) Here is the technique for computing the Mellin transform of $f(x)$.

Answer 2

Let's start with $$ \sum_{n=0}^\infty x^n=\frac1{1-x}\tag{1} $$ Differentiating $(1)$ and multiplying by $x$, we get $$ \sum_{n=0}^\infty nx^n=\frac{x}{(1-x)^2}\tag{2} $$ Taking the odd part of $(2)$ yields $$ \sum_{n=0}^\infty(2n+1)x^{2n+1}=\frac{x(1+x^2)}{(1-x^2)^2}\tag{3} $$ Using $(3)$, we get $$ \begin{align} \sum_{n=0}^\infty\frac{2n+1}{e^{(2n+1)\pi}+1} &=\sum_{n=0}^\infty\sum_{k=1}^\infty(-1)^{k-1}(2n+1)e^{-(2n+1)k\pi}\\ &=\sum_{k=1}^\infty\sum_{n=0}^\infty(-1)^{k-1}(2n+1)e^{-(2n+1)k\pi}\\ &=\sum_{k=1}^\infty(-1)^{k-1}\frac{e^{-k\pi}\left(1+e^{-2k\pi}\right)}{\left(1-e^{-2k\pi}\right)^2}\\ &=\frac12\sum_{k=1}^\infty(-1)^{k-1}\frac{\cosh(k\pi)}{\sinh^2(k\pi)}\tag{4} \end{align} $$

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We can use the formula proven in this answer $$ \pi\cot(\pi z)=\sum_{k\in\mathbb{Z}}\frac1{z+k}\tag{5} $$ to derive $$ \begin{align} \pi\csc(\pi z) &=\pi\cot(\pi z/2)-\pi\cot(\pi z)\\[9pt] &=\sum_{k\in\mathbb{Z}}\frac2{z+2k}-\sum_{k\in\mathbb{Z}}\frac1{z+k}\\ &=\sum_{k\in\mathbb{Z}}\frac{(-1)^k}{z+k}\\ \pi^2\frac{\cos(\pi z)}{\sin^2(\pi z)} &=\sum_{k\in\mathbb{Z}}\frac{(-1)^k}{(z+k)^2}\tag{6} \end{align} $$ then rotate coordinates with $z\mapsto iz$ to get $$ \pi^2\frac{\cosh(\pi z)}{\sinh^2(\pi z)}=\sum_{j\in\mathbb{Z}}\frac{(-1)^j}{(z+ij)^2}\tag{7} $$

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Now plug $(7)$ into $(4)$: $$ \begin{align} \sum_{n=0}^\infty\frac{2n+1}{e^{(2n+1)\pi}+1} &=\frac1{2\pi^2}\sum_{k=1}^\infty\sum_{j\in\mathbb{Z}}(-1)^{j+k-1}\frac1{(k+ij)^2} \\ &=\frac1{2\pi^2}\sum_{k=1}^\infty(-1)^{k-1}\frac1{k^2}\\ &+\frac1{2\pi^2}\sum_{k=1}^\infty\sum_{j=1}^\infty(-1)^{j+k-1}\left(\frac1{(k+ij)^2}+\frac1{(k-ij)^2}\right)\\ &=\frac1{2\pi^2}\frac{\pi^2}{12}\\ &+\frac1{2\pi^2}\sum_{k=1}^\infty\sum_{j=1}^\infty(-1)^{j+k-1}\frac{2(k^2-j^2)}{(k^2+j^2)^2}\\ &=\frac1{24}+0\tag{8} \end{align} $$

Answer 3

The calculation of the Mellin transform of $f(x)$ is not present in the above answer, so I will show it here.

$$\mathfrak{M}\left(\frac{1}{e^{\pi x}+1};s \right) = \int_0^\infty \frac{1}{e^{\pi x}+1} x^{s-1} dx = \int_0^\infty \frac{1}{e^{\pi x}} \frac{1}{1+e^{-\pi x}} x^{s-1} dx \\= \int_0^\infty \frac{1}{e^{\pi x}} \sum_{q\ge 0} (-1)^q e^{-\pi q x} x^{s-1} dx = \int_0^\infty \sum_{q\ge 0} (-1)^q e^{-\pi (q+1) x} x^{s-1} dx \\ = \Gamma(s) \sum_{q\ge 0} (-1)^q \frac{1}{\pi^s (q+1)^s} = \frac{1}{\pi^s} \Gamma(s) (\zeta(s) - 2 \times 2^{-s} \times \zeta(s)) = \frac{1}{\pi^s} \Gamma(s) (1 - 2\times 2^{-s}) \zeta(s).$$ It now follows from the definition of the Mellin transform that $$\mathfrak{M}\left(\frac{x}{e^{\pi x}+1};s \right) = \frac{1}{\pi^{s+1}} \Gamma(s+1) (1 - 2^{-s}) \zeta(s+1).$$

Answer 4

Actually the above is not quite complete, the missing piece is the proof that we can drop the contribution from the pole at $s=-1,$ which is $x/24.$ To verify this we have to show that $$\int_{-i\infty}^{i\infty} \frac{1}{\pi^{s+1}} \Gamma(s+1) (1-2^{-s})^2 \zeta(s+1)\zeta(s) ds = 0.$$ Now from the functional equation of the Riemann Zeta function we see that this integral is equal to $$-\int_{-i\infty}^{i\infty} \frac{\zeta(-s)}{\sin(1/2s\pi)} (2^s-1) (1-2^{-s}) \zeta(s) ds$$ Actually doing the accounting we find that the kernel $$ g(s) = \frac{\zeta(-s)}{\sin(1/2s\pi)} (2^s-1) (1-2^{-s}) \zeta(s) $$ of this integral has the property that $g(s) = - g(-s)$ on the imaginary axis, so the integral is zero.

To see this consider what effect negation has on the individual terms. $$\zeta(-s)\zeta(s) \to \zeta(s)\zeta(-s),$$ $$(2^s-1)(1-2^{-s}) \to (2^{-s}-1)(1-2^s) = (2^s-1)(1-2^{-s}),$$ $$\sin(1/2 s\pi) \to \sin(1/2 (-s)\pi) = -\sin(1/2 s\pi).$$ The first two terms are even and the last one is odd, QED.

Note that we have taken advantage of the fact that $x=1$ ... for other values of $x$ this trick will not go through. Also relevant is that negation (rotation by 180 degrees about the origin) takes the imaginary axis to itself (this is not the case when we are integrating along some other line parallel to the imaginary axis in the right half plane).

Answer 5

We can link the functions $g(x),g(x^{2}),g(x^{4})$ with elliptic integrals $K,E$ and modulus $k$. Also we switch to variable $q$ instead of $x$. The value of $q$ here is $q = e^{-\pi}$ so that $k = 1/\sqrt{2}$.

From this post we can see that \begin{align} g(q) &= \frac{1 - P(q)}{24} = \frac{1}{24}\left\{1 - \left(\frac{2K}{\pi}\right)^{2}\left(\frac{6E}{K} + k^{2} - 5\right)\right\}\tag{1}\\ g(q^{2}) &= \frac{1 - P(q^{2})}{24} = \frac{1}{24}\left\{1 - \left(\frac{2K}{\pi}\right)^{2}\left(\frac{3E}{K} + k^{2} - 2\right)\right\}\tag{2}\\ g(q^{4}) &= \frac{1 - P(q^{4})}{24} = \frac{1}{24}\left\{1 - \left(\frac{2K}{\pi}\right)^{2}\left(\frac{3E}{2K} + \frac{k^{2} - 2}{4}\right)\right\}\tag{3} \end{align} Using the above equations and the fact that $k = 1/\sqrt{2}$ we get $$f(q) = g(q) - 4g(q^{2}) + 4g(q^{4}) = \frac{1}{24}$$ Note: The formula for $P(q^{4})$ is not given in the linked post, but it can be obtained via the same technique as mentioned in the post, namely finding an expression for $\eta(q^{4})$ in terms of $K, k$ and then applying logarithmic differentiation.

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Based on suggestion from robjohn (via comment) I add some details on the theory of elliptic integrals and theta functions so that the identities $(1), (2), (3)$ get some context.

Let us then assume that $k$ is a real number (which is called modulus) with $0 < k < 1$ and let $k' = \sqrt{1 - k^{2}}$ then we define elliptic integrals $$K(k) = \int_{0}^{\pi/2}\frac{dx}{\sqrt{1 - k^{2}\sin^{2}x}}, E(k) = \int_{0}^{\pi/2}\sqrt{1 - k^{2}\sin^{2}x}\,dx\tag{4}$$ The integrals $K(k), E(k), K(k'), E(k')$ are usually denoted by $K, E, K', E'$ when the argument $k, k'$ is available from context. These integrals satisfy the Legendre's Identity $$KE' + K'E - KK' = \frac{\pi}{2}\tag{5}$$ We have the derivatives $$\frac{dK}{dk} = \frac{E - k'^{2}K}{kk'^{2}},\,\,\frac{dE}{dk} = \frac{E - K}{k}\tag{6}$$ It is almost a piece of magic (created by mathemagician Jacobi) that it is possible to obtain the values of $k, k'$ from the values of $K, K'$ using another variable called nome defined by $$q = e^{-\pi K'/K}\tag{7}$$ The desired relations are given by theta functions $\vartheta_{2}(q), \vartheta_{3}(q), \vartheta_{4}(q)$ as follows $$k = \frac{\vartheta_{2}^{2}(q)}{\vartheta_{3}^{2}(q)},\, k' = \frac{\vartheta_{4}^{2}(q)}{\vartheta_{3}^{2}(q)},\, \frac{2K}{\pi} = \vartheta_{3}^{2}(q)\tag{8}$$ where \begin{align} \vartheta_{2}(q) &= \sum_{n = -\infty}^{\infty}q^{(n + 1/2)^{2}} = 2(q^{1/4} + q^{9/4} + q^{25/4} + \cdots )\notag\\ &= 2q^{1/4}\prod_{n = 1}^{\infty}(1 - q^{2n})(1 + q^{2n})^{2}\tag{9a}\\ \vartheta_{3}(q) &= \sum_{n = -\infty}^{\infty}q^{n^{2}} = 1 + 2q + 2q^{4} + 2q^{9} + \cdots\notag\\ &= \prod_{n = 1}^{\infty}(1 - q^{2n})(1 + q^{2n - 1})^{2}\tag{9b}\\ \vartheta_{4}(q) &= \sum_{n = -\infty}^{\infty}(-1)^{n}q^{n^{2}} = 1 - 2q + 2q^{4} - 2q^{9} + \cdots\notag\\ &= \prod_{n = 1}^{\infty}(1 - q^{2n})(1 - q^{2n - 1})^{2}\tag{9c} \end{align} Using the relations $(5), (6), (7)$ it is possible to show that $$\frac{dq}{dk} = \frac{\pi^{2}q}{2kk'^{2}K^{2}}\tag{10}$$ We next introduce the function $\eta(q)$ via $$\eta(q) = q^{1/24}\prod_{n = 1}^{\infty}(1 - q^{n})\tag{11}$$ Logarithmic differentiation of the above relation gives rise to the function $P(q)$ i.e. $$P(q) = 1 - 24\sum_{n = 1}^{\infty}\frac{nq^{n}}{1 - q^{n}} = 24q\frac{d}{dq}\log\eta(q)\tag{12}$$ Using equations $(8), (9a), (9b), (9c)$ (especially the product representations of theta functions) it is possible to express $\eta(q)$ in terms of $k, k', K$ and we have \begin{align} \eta(q) &= 2^{-1/6}\sqrt{\frac{2K}{\pi}}k^{1/12}k'^{1/3}\tag{13a}\\ \eta(q^{2}) &= 2^{-1/3}\sqrt{\frac{2K}{\pi}}(kk')^{1/6}\tag{13b}\\ \eta(q^{4}) &= 2^{-2/3}\sqrt{\frac{2K}{\pi}}k^{1/3}k'^{1/12}\tag{13c} \end{align} I obtained the expression of $\eta(q^{2})$ directly using the product representation of theta functions as $$2\{\eta(q^{2})\}^{3} = \vartheta_{2}(q)\vartheta_{3}(q)\vartheta_{4}(q)$$ and then equation $(13b)$ follows by using equation $(8)$. The expressions for $\eta(q), \eta(q^{4})$ were obtained by applying Landen's Transformation on these expressions (thus if we replace $k$ by $2\sqrt{k}/(1 + k)$ and $K$ by $(1 + k)K$ the variable $q^{2}$ is replaced by $q$ and if we replace $k$ by $(1 - k')/(1 + k')$ and $K$ by $(1 + k')K/2$ the variable $q$ is replaced by $q^{2}$). The identities $(1), (2)$ and $(3)$ of my answer are obtained by logarithmic differentiation of the above expressions for $\eta(q), \eta(q^{2}), \eta(q^{4})$.

Answer 6

$$S=\sum_{n=1}^{\infty }\frac{2n-1}{1+e^{\pi (2n-1)}},$$ first we know that $$\frac{1}{1+e^x}=\sum_{k=1}^{\infty }(-1)^{k-1}e^{-kx}$$ therefore $$S=\sum_{n=1}^{\infty }(2n-1)\sum_{k=1}^{\infty }(-1)^{k-1}e^{-k(2n-1)\pi }=\sum_{k=1}^{\infty }(-1)^{k-1}\sum_{n=1}^{\infty }(2n-1)e^{-k(2n-1)\pi }$$

but we know $$ \sum_{n=1}^{\infty }e^{-(2n-1)\pi x}=\frac{1}{2\sinh(\pi x)},$$ for all $x>0$ but $$\frac{1}{2\sinh(\pi x)}=(\frac{i}{2\pi })(\frac{\pi }{\sin(i\pi x)})$$ and $$\int_{0}^{\infty }\frac{t^{-x}}{1+t}dt=\frac{\pi }{\sin(\pi x)}$$ therefore $$\int_{0}^{\infty }\frac{t^{-x}}{1+t}dt=\int_{0}^{1}\frac{t^{-x}}{1+t}dt+\int_{0}^{1}\frac{t^{x-1}}{1+t}dt=\int_{0}^{1}\frac{t^{-x}+t^{x-1}}{1+t}dt\\ \\ \frac{\pi }{\sin(\pi x)}=\int_{0}^{1}\frac{t^{-x}+t^{x-1}}{1+t}dt=\sum_{k=1}^{\infty }(-1)^{k-1}(\frac{1}{k-x}+\frac{1}{k-1+x})\\ \therefore from\ \frac{\pi }{\sin(i\pi x)}=\int_{0}^{\infty }\frac{t^{-ix}+t^{ix-1}}{1+t}dt=\sum_{k=1}^{\infty }(-1)^{k-1}(\frac{1}{k-ix}+\frac{1}{k-1+ix})\\ \\ =\sum_{k=1}^{\infty }(-1)^{k-1}(\frac{k+ix}{x^2+k^2}+\frac{k-1-ix}{(k-1^2)+x^2})=\sum_{k=1}^{\infty }(-1)^{k-1}(\frac{k}{k^2+x^2}+\frac{k-1}{x^2+(k-1)^2})\\ \\ +i\sum_{k=1}^{\infty }(-1)^{k-1}(\frac{x}{x^2+k^2}-\frac{x}{x^2+(k-1)^2})=i(\sum_{k=1}^{\infty }(-1)^{k-1}\frac{x}{x^2+k^2}-\sum_{k=1}^{\infty }(-1)^{k}\frac{x}{x^2+k^2})\\ \\ \\ =i[\frac{-1}{x}+\sum_{k=1}^{\infty }(-1)^{k-1}\frac{x}{x^2+k^2}+\sum_{k=1}^{\infty }(-1)^{k-1}\frac{x}{x^2+k^2}]=i[\frac{-1}{x}+2\sum_{k=1}^{\infty }(-1)^{k-1}\frac{x}{x^2+k^2}]\\ \\\therefore \frac{\pi }{\sinh(\pi x)}=\frac{1}{x}-2\sum_{k=1}^{\infty }(-1)^{k-1}\frac{x}{k^2+x^2}=\int_{0}^{1 }\frac{t^{-ix}+t^{ix-1}}{1+t}dt$$

Answer 7

Integrate the function $$f(z) = \frac{z e^{iz}}{\cosh (z) + \cos(z)} \, $$ around the contour $[-\sqrt{2} \pi N, \sqrt{2} \pi N] \cup \sqrt{2} \pi Ne^{i[0, \pi]}$, where $N$ is a positive integer.

For every positive integer $N$, the contour passes halfway between adjacent poles of $f(z)$.

The integral vanishes on the semicircle as $N \to \infty$. This is because $|\cosh(z)|$ growns exponentially as $\Re(z) \to \pm \infty$, while $|e^{iz}|$ decays exponentially to zero as $\Im(z) \to + \infty$.

We therefore have

$\begin{align} \int_{-\infty}^{\infty}f(x) \, \mathrm dx &= 2 \pi i \left(\sum_{n=0}^{\infty}\operatorname{Res} \left[f(z), \frac{(2n+1) \pi (1+i)}{2} \right] + \sum_{n=0}^{\infty}\operatorname{Res} \left[f(z), \frac{(2n+1) \pi (-1+i)}{2} \right]\right) \\ &= 2 \pi i \left(\sum_{n=0}^{\infty} \lim_{z \to \frac{(2n+1) \pi (1+i)}{2} }\frac{ze^{iz}}{\sinh(z) - \sin(z)} + \sum_{n=0}^{\infty} \lim_{z \to \frac{(2n+1) \pi (-1+i)}{2} }\frac{ze^{iz}}{\sinh(z) - \sin(z)} \right) \\ &= 2 \pi i \left(\frac{\pi}{2}\sum_{n=0}^{\infty} \frac{(2n+1)e^{-(2n+1) \pi /2}}{\cosh \left(\frac{(2n+1) \pi }{2} \right)} + \frac{\pi}{2}\sum_{n=0}^{\infty} \frac{(2n+1)e^{-(2n+1) \pi /2}}{\cosh \left(\frac{(2n+1) \pi }{2} \right)}\right) \\ &= 2 \pi^{2} i \sum_{n=0}^{\infty} \frac{(2n+1) e^{- (2n+1) \pi /2}}{\cosh \left(\frac{(2n+1)\pi }{2} \right)} \\ &= 4 \pi^{2}i \sum_{n=0}^{\infty} \frac{2n+1}{e^{(2n+1)\pi }+1}. \end{align}$

Equating the imaginary parts on both sides of the equation, we get $$ \begin{align} \sum_{n=0}^{\infty} \frac{2n+1}{e^{(2n+1) \pi }+1} &= \frac{1}{4 \pi^{2}} \int_{-\infty}^{\infty} \frac{x \sin(x)}{\cosh (x) + \cos(x)}\, \mathrm dx \\ &= \frac{1}{2 \pi^{2}}\int_{0}^{\infty} \frac{x \sin(x)}{\cosh (x) + \cos(x)} \, \mathrm dx \\ &= \frac{1}{\pi^{2}} \, \Im \int_{0}^{\infty} x \sum_{n=1}^{\infty} (-1)^{n-1} e^{(-1+i)nx} \, \mathrm dx \\ &= \frac{1}{\pi^{2}} \, \Im \sum_{n=1}^{\infty} (-1)^{n-1} \int_{0}^{\infty} x e^{(-1+i)nx} \, \mathrm dx \\ &= \frac{1}{\pi^{2}} \, \Im \sum_{n=1}^{\infty} (-1)^{n-1} \frac{1}{(1-i)^{2}n^{2}} \\ &= \frac{1}{2 \pi^{2}} \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^{2}} \\ &= \frac{1}{24} . \end{align} $$