Limit approach to finding $1+2+3+4+\ldots$

Question

When exploring the divergent series consisting of the sum of all natural numbers

$$\sum_{k=1}^\infty k=1+2+3+4+\ldots$$

I came across the following identity involving a one-sided limit:

$$\lim_{x\to0^-}\sum_{k=1}^\infty k\exp(kx)\cos(kx)=-\frac{1}{12}$$

Using zeta function regularization yields the same value:

$$\sum_{k=1}^\infty k^{-s}=\zeta(s)$$ $$\zeta(-1)=-\frac{1}{12}$$

In general, I found that for $y\neq0$ and $n=1,5,9,13,\ldots$ (i.e. $4m+1$ where $m\in\mathbb{N}$),

$$\lim_{x\to0^-}\sum_{k=1}^\infty k^n\exp(kxy)\cos(kxy)=\zeta(-n)$$

However, a similar limit did not exist for other powers of $k$, e.g.

$$\lim_{x\to 0^-}\sum_{k=1}^\infty k^2\exp(kx)\cos(kx)$$ $$\lim_{x\to 0^-}\sum_{k=1}^\infty k^3\exp(kx)\cos(kx)$$

The regularized values of the corresponding series are $\zeta(-2)=0$ and $\zeta(-3)=\frac{1}{120}$.

Given this information, I have the following questions:

1. What is the connection between the limit approach and the zeta function approach?
2. Why does the limit expression only seem to converge for $n=4m+1$?
3. Can the limit approach be used to find the sum for other powers of $k$? If so, how?

Here is a plot I made with Mathematica using the command

Plot[Evaluate[Sum[k*Exp[x*k]*Cos[x*k], {k, 1, Infinity}]], {x, -16, 16}, PlotRange -> {-.25, .25}, AspectRatio -> 1]

Notice how it approaches $-1/12\approx-0.08333$ as $x$ approaches $0$.

Further information:

$$\sum_{k=1}^\infty k^n\exp(kx)=\mathrm{Li}_{-n}(\exp(x))=\frac{n!}{(-x)^{n+1}}+\zeta(-n)+O(x)$$

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Edit:

Based on Micah's answer, we seek an $f$ such that $f(s,0) = 1$ and \begin{align} \int_0^\infty x^s f(s,x) \,\mathrm{d}x &= 0 \end{align}

Let \begin{align} f(s,x) &= \mathrm{e}^{-x}(1+ax) \end{align}

Then $f(s,0) = 1$. Assuming $\mathrm{Re}(s) > -1$, \begin{align} 0 &= \int_0^\infty x^s f(s,x) \,\mathrm{d}x \\ &= \int_0^\infty x^s \mathrm{e}^{-x}(1+ax) \,\mathrm{d}x \\ &= \int_0^\infty x^s \mathrm{e}^{-x} \,\mathrm{d}x + a \int_0^\infty x^{s+1} \mathrm{e}^{-x} \\ &= \Gamma(s+1) + a (s+1) \Gamma(s+1) \\ &= (1 + a(s+1)) \Gamma(s+1) \\ \Longrightarrow a &= -\frac{1}{s+1} \\ \Longrightarrow f(s,x) &= \mathrm{e}^{-x}\left( 1 - \frac{x}{s+1} \right) \end{align}

Thus \begin{align} \zeta(-s) &= \lim_{\varepsilon \rightarrow 0^+} \lim_{m \rightarrow \infty} \sum_{n=1}^m n^s f(s, n \varepsilon) \\ &\overset{\star}{=} \lim_{m \rightarrow \infty} \lim_{\varepsilon \rightarrow 0^+} \sum_{n=1}^m n^s f(s, n \varepsilon) \\ &= \lim_{m \rightarrow \infty} \sum_{n=1}^m n^s f(s, 0) \\ &= \lim_{m \rightarrow \infty} \sum_{n=1}^m n^s \\ &= \sum_{n=1}^\infty n^s \end{align}

where the star indicates the non-rigorous step of exchanging limits. In fact, this regulator seems to work for all $s \neq -1$ (see some plots for $s < -1$ below). Hence we can also regularize the harmonic series as follows: \begin{align} \gamma &= \frac{1}{2} \lim_{\varepsilon \rightarrow 0^+} (\zeta(1+\varepsilon) + \zeta(1-\varepsilon)) \\ &= \frac{1}{2} \lim_{\varepsilon \rightarrow 0^+} \left(\lim_{m \rightarrow \infty} \sum_{n=1}^m n^{-1-\varepsilon} f(-1-\varepsilon, n \varepsilon) + \lim_{m \rightarrow \infty} \sum_{n=1}^m n^{-1+\varepsilon} f(-1+\varepsilon, n \varepsilon) \right) \\ &= \frac{1}{2} \lim_{\varepsilon \rightarrow 0^+} \lim_{m \rightarrow \infty} \left(\sum_{n=1}^m n^{-1-\varepsilon} f(-1-\varepsilon, n \varepsilon) + \sum_{n=1}^m n^{-1+\varepsilon} f(-1+\varepsilon, n \varepsilon) \right) \\ &\overset{\star}{=} \frac{1}{2} \lim_{m \rightarrow \infty} \lim_{\varepsilon \rightarrow 0^+} \left(\sum_{n=1}^m n^{-1-\varepsilon} f(-1-\varepsilon, n \varepsilon) + \sum_{n=1}^m n^{-1+\varepsilon} f(-1+\varepsilon, n \varepsilon) \right) \\ &= \frac{1}{2} \lim_{m \rightarrow \infty} \left(\sum_{n=1}^m n^{-1} f(-1, 0) + \sum_{n=1}^m n^{-1} f(-1, 0) \right) \\ &= \frac{1}{2} \lim_{m \rightarrow \infty} \left(\sum_{n=1}^m n^{-1} + \sum_{n=1}^m n^{-1} \right) \\ &= \lim_{m \rightarrow \infty} \sum_{n=1}^m n^{-1} \\ &= \sum_{n=1}^\infty n^{-1} \\ \end{align}

Here is the Mathematica code and its plots:

f[s_, x_] := Exp[-x] (1 + a x)
g[s_, t_] := 
 Evaluate@Simplify[
   f[s, t] /. Solve[Integrate[x^s f[s, x], {x, 0, Infinity}] == 0, a],
    Assumptions -> Re[s] > -1]
g[s, t]
Table[{s, 
    Plot[{Zeta[-s], 
      Sum[n^s g[s, n \[Epsilon]], {n, 1, 1000}]}, {\[Epsilon], 0, 1}, 
     Evaluated -> True]}, {s, -4, 4, 1/2}] // TableForm // Quiet
Plot[{EulerGamma, 
  Sum[(n^(-1 + \[Epsilon]) g[-1 + \[Epsilon], n \[Epsilon]] + 
    n^(-1 - \[Epsilon]) g[-1 - \[Epsilon], n \[Epsilon]])/
   2, {n, 1, 1000}]}, {\[Epsilon], 0, 1}, Evaluated -> True]

Answer 1

In Terence Tao's blog post about relating divergent series to negative zeta values, he uses Euler-Maclaurin summation to show that if $\eta$ is smooth and compactly-supported, and $\eta(0)=1$, then $$ \sum_{n=0}^\infty n^s \eta(n/N)=C_{\eta,s} N^{s+1} + \zeta(-s) + O(1/N) $$ for any positive integer $s$, where $C_{\eta,s}=\int_0^\infty x^s \eta(x) \, dx$. In fact, it's possible to extend his proof to the case where $\eta$ is a Schwartz function — see below for a proof.

In particular, consider $\eta(x)=e^{-x}\cos x$. After a change of variables we can write your sum in the above form, with this $\eta$. Tao discusses the connection between this asymptotic form and the analytic continuation of the zeta function; I haven't gone through that part of his blog post in detail, but assuming that it, too, could be made to work when $\eta$ was Schwartz-class, that would presumably answer your first question about the connection between your sum and the zeta-regularized form of the original divergent sum.

To answer your second question, we can look at the leading coefficient $C_{\eta,s}=\int_0^\infty x^s e^{-x} \cos x \, dx$. After a bunch of annoying but straightforward integration by parts, it's possible to compute that $\int_0^\infty x^s e^{-x} \cos x=0$ when $s \equiv 1 \pmod{4}$, and $\int_0^\infty x^s e^{-x} \sin x=0$ when $s \equiv 3 \pmod{4}$. That is, whenever $s \equiv 1 \pmod{4}$, the $\eta(x)$ you've chosen coincidentally causes the leading term of the asymptotic expansion to drop out, exposing the zeta value which is the constant term of the expansion.

You can expose $\zeta(-4k)$ in a similar fashion by taking $\eta(x)=e^{-x} \frac{\sin x}{x}$, though since it's zero, that's not particularly interesting. For example, here's $\zeta(-4)$. Similarly, you can expose $\zeta(-4k+2)$ by taking $\eta(x)=e^{-x}(\cos x + \sin x)$. The interesting question — to which I don't know the answer — would be whether there's a nice choice of $\eta$ that exposes $\zeta(-4k+1)$, since that's the other case where it's actually nonzero. But this provides at least a partial answer to your third question.

Also note that if we take $\eta(x)=e^{-x}$, Tao's formula gives your "further information" expansion, as $\int_0^\infty x^n e^{-x} \, dx = n!$.

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To prove Tao's formula whenever $\eta$ is smooth and rapidly decreasing, we'll use the infinite-series version of the Euler-Maclaurin formula (as found, e.g., in these notes): if $f(t)$ and all its derivatives tend to zero as $t \to \infty$, then $$ \sum_{n=0}^\infty f(n) = \int_{0}^\infty f(t) \, dt + \frac{1}{2} f(0) - \sum_{i=2}^k \frac{b_i}{i!} f^{(i-1)}(0)-\int_0^\infty \frac{B_k(\{1-t\})}{k!} f^{(k)}(t) \, dt $$ for any fixed integer $k$, where the $b_k$ are the Bernoulli numbers, the $B_k$ are the Bernoulli polynomials, and $\{x\}$ denotes the fractional part of $x$.

In our case, we set $f(t)=t^s \eta(t/N)$ and use the $(s+2)$nd-order expansion. Taking each term in the formula one-by-one, note that:

$$ \int_0^\infty f(t) \, dt=\int_0^\infty t^s \eta(t/N) \, dt = N^{s+1}\int_0^\infty x^s \eta(x) \, dx=C_{\eta,s}N^{s+1} $$ after making the substitution $x=t/N$; $$ \frac{1}{2}f(0)=0 $$ is immediate; \begin{align} -\sum_{i=2}^{s+2} \frac{b_i}{i!} f^{(i-1)}(0)&=-\frac{b_{s+1}}{(s+1)!} s! \eta(0)-\frac{b_{s+2}}{(s+2)!}\left(\frac{(s+1)!}{N} \eta'(0)\right)\\&=-\frac{b_{s+1}}{s+1} + O(1/N)\\&=\zeta(-s) + O(1/N) \end{align} as the first $s-1$ derivatives of $f$ vanish, and \begin{align} -\int_0^\infty \frac{B_{s+2}(\{1-t\})}{(s+2)!} f^{(s+2)}(t) \, dt &= -\int_0^\infty \frac{B_{s+2}(\{1-t\})}{(s+2)!} \frac{d^{s+2}}{dt^{s+2}}\left(t^s \eta(t/N)\right) \\ &=-\int_0^\infty \frac{B_{s+2}(\{1-Nx\})}{(s+2)!} \frac{1}{N^{s+2}}\frac{d^{s+2}}{dx^{s+2}}\left(N^sx^s \eta(x)\right) N \, dx \\ &=-\frac{1}{N} \int_0^\infty \frac{B_{s+2}(\{1-Nx\})}{(s+2)!} \frac{d^{s+2}}{dx^{s+2}}\left(x^s \eta(x)\right) \, dx \end{align} where we again make the substitution $x=t/N$.

This last quantity is bounded in magnitude by $\frac{1}{N}\frac{M}{(s+2)!}\int_0^\infty \frac{d^{s+2}}{dx^{s+2}}\left(x^s \eta(x)\right) \, dx$, where $M$ is the maximum value of $B_{n+2}(x)$ on $[0,1]$. Thus it too is $O(1/N)$ so long as that integral converges.

So Tao's formula works, not only for compactly supported $\eta$, but for $\eta$ such that $f(t)=t^s \eta(t)$ as well as all of $f$'s derivatives are integrable on $[0,\infty)$. In particular, it will work for all $s$ whenever $\eta$ is Schwartz on $[0,\infty)$.

Answer 2

The infamous $-\frac{1}{12}$ pops up as the coefficient of of a range of expansions, such as

$\dfrac{z}{{\mathrm e}^{-z}-1}=-1-\dfrac{1}{2}z-\dfrac{1}{12}z^2+{\mathcal O}(z^3)$

like in the Todd class or similarly in the Baker–Campbell–Hausdorff formula and so on. In your post, you are led to it by ${\mathrm e}^{kx}\cos(kx)$, which after the reparametrization $x\mapsto \log(z)$ I'd write as as $z^k\cos(\log(z^k))$. But in any case, the higher order expansion terms of the cosine make this a not so nice choice and you get a lot of cases where the limit procedure doesn't work. I've worked out some broader perspectives..

The analytical continuation, e.g. involved in defining the Zeta function*

$\zeta(s) = \dfrac{1}{\Gamma(s)} \int_{0}^{\infty} \dfrac{x ^ {s-1}}{{\mathrm e} ^ x - 1}\mathrm{d}x,$

is a process which smoothly probes local values of the functional expressions $f$ in $\sum_{k=0}^\infty f(k)$. You can formulate the regularization of the sum in a way that reflects that, e.g. by using a local mean like

$\langle f(k)\rangle:=\int_{k}^{k+1}f(k')\,{\mathrm d}k'$.

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So the limit $\lim_{z\to 1}$ of the sum

$0+1\,z^1+2\,z^2+3\,z^3+\dots$

diverges, because of the pole in

$\sum_{k=0}^\infty k\,z^k=z\dfrac{{\mathrm d}}{{\mathrm d}z}\sum_{k=0}^\infty z^k=z\dfrac{{\mathrm d}}{{\mathrm d}z}\dfrac{1}{1-z}=\dfrac{z}{(z-1)^2}, \hspace{1cm} z\in(0,1)$

So let's consider the sum of smooth deviations. With

$\langle k\,z^k\rangle=z\dfrac{{\mathrm d}}{{\mathrm d}z}\langle z^k\rangle=z\dfrac{{\mathrm d}}{{\mathrm d}z}\langle {\mathrm e}^{k \log(z)}\rangle=\left.z\dfrac{{\mathrm d}}{{\mathrm d}z}\dfrac{z^{k'}}{\log(z)}\right|_{k}^{k+1}$.

we find the sum $\sum_{k=0}^n\langle k\,z^k\rangle$ involves canceling upper and lower bounds and we're left with $\dfrac{z^0}{\log(z)^2}$ plus terms suppressed by $z^n$. Finally, using the expansion

$r^2\dfrac{1}{\log(1+r)^2}=\dfrac{1}{1-r+\left(1-\frac{1}{1!\,2!\,3!}\right)r^2+{\mathrm{O}}(r^3)}=1+r+\dfrac{1}{1!\,2!\,3!}r^2+{\mathrm{O}}(r^3),$

we find

$\sum_{k=0}^\infty \left(k\,z^k-\langle k\,z^k\rangle\right)=\dfrac{z}{(z-1)^2}-\dfrac{1}{\log(z)^2}=-\dfrac{1}{12}+{\mathcal O}\left((z-1)^1\right).$

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The picture shows the two functions $\dfrac{z}{(z-1)^2}$ and $\dfrac{1}{\log(z)^2}$, as well as their difference (blue, red, yellow). While the functions themselves clearly have a pole at $z=1$, their difference converges against

$$-\frac{1}{1!\,2!\,3!}=-\frac{1}{12}=-0.08{\dot 3}.$$

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And here's the Mathematica calculation for higher $n$, as well as the corresponding values of the Zeta function (which match)

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I've tried to generalize those in various directions and stumbled upon some futher relations. For example, when comparing finite differences to their first order approximations, we find

$\dfrac{f'(x)\,h}{f(x+h)-f(x)}=1-\dfrac{f''(x)}{2!}\left(\dfrac{h}{f'(x)}\right)+\left(\dfrac{f''(x)\,f''(x)}{2!\,2!}-\dfrac{f'(x)\,f'''(x)}{1!\,3!}\right)\left(\dfrac{h}{f'(x)}\right)^2+{\mathcal O}(h^3).$

Note that $\dfrac{1}{2!\,2!}-\dfrac{1}{1!\,3!}=\dfrac{1}{2!\,3!}(3-2)=\dfrac{1}{12}$, which is the coeffient $\dfrac{1}{2!}B_2$ in the related MacLaurin formula

And the log-substraction scheme can be performed more generally for higher order poles $\frac{1}{(z-1)^n}$ using

$\dfrac{1}{\log(z)^n}=\dfrac{1}{(z-1)^n}\left(1+\frac{n}{2}(z-1)+\frac{n}{2}\frac{3n-5}{12}(z-1)^2+\frac{n}{2}\frac{(n-2)(n-3)}{24}(z-1)^3+\dots\right)$

Here, e.g. plug in $n=2$ to get $\dfrac{n}{2}\dfrac{3n-5}{12}(z-1)^2=\dfrac{1}{12}$.

*As I see from your profile you're interested a physics (I'm a physicist myself) I like to point out how the body of the integrand in that Zeta function representation is the expression of Planck's law :). The exponential function ${\mathrm e}^x\cdot{\mathrm e}^y={\mathrm e}^{x+y}$, say in statistical physics where $-\frac{1}{12}$ come to play a role, enters here because the probabilities of divided systems behave multiplicative, while the energy, by definition, behaves additive.

Answer 3

Explanation for the Behavior of the Series (Parts $2$ and $3$)

Note that $$ \begin{align} \sum_{k=0}^\infty e^{ikx} &=\frac1{1-e^{ix}}\\ &=\frac12+\frac i2\cot\left(\frac x2\right)\\ &=\frac ix+\frac12-i\,\underbrace{\left(\frac1x-\frac12\cot\left(\frac x2\right)\right)}_{\substack{\text{odd function of $x$}\\\text{analytic on $(-\pi,\pi)$}}}\tag{1} \end{align} $$

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Odd Exponents

Taking $2n-1$ derivatives of $(1)$, we get $$ \sum_{k=0}^\infty k^{2n-1}e^{ikx}=\frac{(-1)^n(2n-1)!}{x^{2n}}+(-1)^n\underbrace{\frac{\mathrm{d}^{2n-1}}{\mathrm{d}x^{2n-1}}\left(\frac1x-\frac12\cot\left(\frac x2\right)\right)}_{\substack{\text{even function of $x$}\\\text{analytic on $(-\pi,\pi)$}}}\tag{2} $$

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Exponents which are $\boldsymbol{1\bmod{4}}$

If $n$ is odd, so that $2n-1\equiv1\pmod{4}$, substitute $x=(1-i)y$ in $(2)$ to get $$ \sum_{k=0}^\infty k^{2n-1}e^{k(1+i)y} =\underbrace{\frac{(2n-1)!}{(2i)^ny^{2n}}}_{\substack{\text{pure imaginary}\\\text{since $n$ is odd}}} -\underbrace{\frac{\mathrm{d}^{2n-1}}{\mathrm{d}x^{2n-1}}\left(\frac1x-\frac12\cot\left(\frac x2\right)\right)}_{\text{plug in $x=(1-i)y$}}\tag{3} $$ Thus, the real part of $(3)$ is $$ \sum_{k=0}^\infty k^{2n-1}e^{ky}\cos(ky)=\left.-\frac{\mathrm{d}^{2n-1}}{\mathrm{d}x^{2n-1}}\left(\frac1x-\frac12\cot\left(\frac x2\right)\right)\right|_{x=(1-i)y}\tag{4} $$ Taking the limit of $(4)$ as $y\to0$ yields the identity given in the question $$ \begin{align} &\text{When $n$ is odd; that is, }2n-1\equiv1\pmod{4}:\\ &\bbox[5px,border:2px solid #C0A000]{\lim_{y\to0^-}\sum_{k=0}^\infty k^{2n-1}e^{ky}\cos(ky)=\left.-\frac{\mathrm{d}^{2n-1}}{\mathrm{d}x^{2n-1}}\left(\frac1x-\frac12\cot\left(\frac x2\right)\right)\right|_{x=0}}\tag{5} \end{align} $$

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Exponents which are $\boldsymbol{3\bmod{4}}$

If $n$ is even, so that $2n-1\equiv3\pmod{4}$, the situation is a bit more complicated. We can still substitute $x=(1-i)y$ in $(2)$, getting $$ \sum_{k=0}^\infty k^{2n-1}e^{k(1+i)y} =\underbrace{\frac{(2n-1)!}{(2i)^ny^{2n}}}_{\text{real since $n$ is even}} +\underbrace{\frac{\mathrm{d}^{2n-1}}{\mathrm{d}x^{2n-1}}\left(\frac1x-\frac12\cot\left(\frac x2\right)\right)}_{\text{plug in $x=(1-i)y$}}\tag{6} $$ We can also substitute $x=-iy$ in $(2)$, getting $$ \sum_{k=0}^\infty k^{2n-1}e^{ky}=\frac{(2n-1)!}{y^{2n}}+\underbrace{\frac{\mathrm{d}^{2n-1}}{\mathrm{d}x^{2n-1}}\left(\frac1x-\frac12\cot\left(\frac x2\right)\right)}_{\text{plug in $x=-iy$}}\tag{7} $$ Taking a linear combination of $(6)$ and $(7)$ to cancel the singular part and maintain the constant part, we get $$ \begin{align} &\text{When $n$ is even; that is, }2n-1\equiv3\pmod{4}:\\ &\bbox[5px,border:2px solid #C0A000]{\lim_{y\to0^-}\sum_{k=0}^\infty k^{2n-1}e^{ky}\frac{2^n\cos(ky)-(-1)^{n/2}}{2^n-(-1)^{n/2}} =\left.\frac{\mathrm{d}^{2n-1}}{\mathrm{d}x^{2n-1}}\left(\frac1x-\frac12\cot\left(\frac x2\right)\right)\right|_{x=0}}\tag{8} \end{align} $$

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Even Exponents

Taking $2n$ derivatives of $(1)$, we get $$ \sum_{k=0}^\infty k^{2n}e^{ikx} =(-1)^n\frac{(2n)!}{x^{2n+1}}i -(-1)^ni\underbrace{\frac{\mathrm{d}^{2n}}{\mathrm{d}x^{2n}}\left(\frac1x-\frac12\cot\left(\frac x2\right)\right)}_{\substack{\text{odd function of $x$}\\\text{analytic on $(-\pi,\pi)$}}}\tag{9} $$ Substitute $x=(1-i)y$ in $(9)$ to get $$ \sum_{k=0}^\infty k^{2n}e^{k(1+i)y} =-\frac{(2n)!}{y^{2n+1}}\frac{1+i}{(2i)^{n+1}} -(-1)^ni\underbrace{\frac{\mathrm{d}^{2n}}{\mathrm{d}x^{2n}}\left(\frac1x-\frac12\cot\left(\frac x2\right)\right)}_{\text{plug in $x=(1-i)y$}}\tag{10} $$ We can also substitute $x=-iy$ in $(9)$ to get $$ \sum_{k=0}^\infty k^{2n}e^{ky} =-\frac{(2n)!}{y^{2n+1}} -(-1)^ni\underbrace{\frac{\mathrm{d}^{2n}}{\mathrm{d}x^{2n}}\left(\frac1x-\frac12\cot\left(\frac x2\right)\right)}_{\text{plug in $x=-iy$}}\tag{11} $$ Taking a linear combination of $(10)$ and $(11)$ to cancel the singular part and maintain the constant part, we get $$ \bbox[5px,border:2px solid #C0A000]{\lim_{y\to0^-}\sum_{k=0}^\infty k^{2n}e^{ky}\frac{2^{n+1}\cos(ky)+(-1)^{\lfloor(n-1)/2\rfloor}}{2^{n+1}+(-1)^{\lfloor(n-1)/2\rfloor}}=0}\tag{12} $$ where $(-1)^{\lfloor(n-1)/2\rfloor}=\mathrm{Re}\left(\frac{i-1}{i^n}\right)$.

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Unified Expression

Combining $(5)$, $(8)$, and $(12)$ yields $$ \bbox[5px,border:2px solid #C0A000]{\lim_{x\to0^-}\sum_{k=0}^\infty k^ne^{kx}\frac{2^{\lceil n/2\rceil}\cos(kx)+c_n}{2^{\lceil n/2\rceil}+c_n}=\zeta(-n)}\tag{13} $$ where $c_n$ is given by $$ c_n=-\tfrac14(-1)^{\lceil n/4\rceil}\left(2+(-1)^{\lceil n/2\rceil}\left(1-(-1)^n\right)\right) $$ or by $$ c_n=-(-1)^{\lceil n/4\rceil}\sin^2\left(\tfrac\pi4(n-1)\right) $$ or in the following table $$ \small\begin{array}{c|r|r|r|r|r|r|r} n\bmod8&1&2&3&4&5&6&7&8\\ \hline c_n&\hphantom{-}0&\hphantom{-}\frac12&\hphantom{-}1&\hphantom{-}\frac12&\hphantom{-}0&-\frac12&-1&-\frac12 \end{array}\tag{14} $$