An advanced problem one of my friends gave me which I think would be interesting to discuss: \[ \text {Compute } \displaystyle\sum_{n=1}^{\infty} \dfrac {1}{n^2+1}. \]

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TopNewestLet \(f(x) = \cosh x\), and let's try to find the Fourier Series of \(f(x)\) on \([-\pi,\pi]\).

\(c_n = \dfrac{1}{2\pi}\displaystyle\int_{-\pi}^{\pi}e^{-inx}\cosh x\,dx\) \(= \dfrac{1}{4\pi}\displaystyle\int_{-\pi}^{\pi}e^{-inx}(e^x+e^{-x})\,dx\) \(= \dfrac{1}{4\pi}\displaystyle\int_{-\pi}^{\pi}(e^{(1-in)x}+e^{(-1-in)x})\,dx\)

\(= \dfrac{1}{4\pi}\left[\dfrac{e^{(1-in)x}}{1-in} + \dfrac{e^{(-1-in)x}}{-1-in}\right]_{-\pi}^{\pi}\) \(= \dfrac{1}{4\pi}\left[\dfrac{e^{(1-in)\pi} - e^{-(1-in)\pi}}{1-in} + \dfrac{e^{(-1-in)\pi} - e^{-(-1-in)\pi}}{-1-in}\right]\)

\(= \dfrac{(-1)^n}{4\pi}\left[\dfrac{e^{\pi} - e^{-\pi}}{1-in} + \dfrac{e^{-\pi} - e^{\pi}}{-1-in}\right]\) \(= \dfrac{(-1)^n\sinh \pi}{2\pi}\left[\dfrac{1}{1-in} - \dfrac{1}{1+in}\right]\) \(= \dfrac{(-1)^n\sinh \pi}{\pi(n^2+1)}\).

Therefore, \(f(x) = \cosh x = \displaystyle\sum_{n = -\infty}^{\infty}\dfrac{(-1)^n\sinh \pi}{\pi(n^2+1)}e^{inx}\).

Thus, \(f(\pi) = \cosh \pi = \displaystyle\sum_{n = -\infty}^{\infty}\dfrac{\sinh \pi}{\pi(n^2+1)}\), and so, \(\displaystyle\sum_{n = -\infty}^{\infty}\dfrac{1}{n^2+1} = \dfrac{\pi \cosh \pi}{\sinh \pi}\).

Subtracting the \(n = 0\) term and dividing by \(2\) gives: \(\displaystyle\sum_{n = 1}^{\infty}\dfrac{1}{n^2+1} = \dfrac{\pi \cosh \pi}{2\sinh \pi} - \dfrac{1}{2} = \dfrac{\pi}{2}\coth \pi - \dfrac{1}{2}\).

EDIT: This is essentially what C L. did, except with complex Fourier coefficients instead. – Jimmy Kariznov · 4 years ago

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– C Lim · 4 years ago

We both posted the exact same solution at the same time. :DLog in to reply

Bob's right: the answer reminded me of some Fourier transform computations I did some years back.

Let \(f(x) = \cosh(x)\) on the interval \( (-\pi, +\pi] \) and repeated periodically through the real line via \(f(x+2\pi) = f(x)\). It's continuous and nice enough that its Fourier series converges pointwise.

Convergence of Fourier TransformSpecifically, at the point \(x = \pi\), the one-sided derivatives exist:

\[ \lim_{x \to \pi^-} \frac{f(x) - \cosh\pi} {x - \pi} = \sinh\pi, \quad \lim_{x\to \pi^+} \frac{f(x) - \cosh\pi} {x - \pi} = \sinh(-\pi) = -\sinh \pi\]

where the second equality follows from \( f(x) = \cosh(x - 2\pi) \) when x is slightly more than \(2 \pi\). Furthermore, the one-sided derivatives of \(f(x)\) have only discrete discontinuities. One can then apply Dini criterion to show that the Fourier series converges.

Computing the Fourier TransformOmitting the details, the Fourier coefficients of \( f(x) \) are given by:

\[ \begin{align*} a_n &= \frac 1 \pi \int_{-\pi}^{\pi} f(x)\cos nx dx = \frac 1 \pi \int_{-\pi}^\pi \cosh x \cos nx dx = 2(-1)^n \frac{\sinh \pi}{\pi},\\ b_n &= \frac 1 \pi \int_{-\pi}^{\pi} f(x)\sin nx dx = \frac 1 \pi \int_{-\pi}^\pi \cosh x \sin nx dx = 0.\end{align*} \]

Thus, the Fourier expansion gives:

\[ \begin{align*} f(x) &= \frac 1 2 a_0 + \sum_{n=1}^\infty (a_n \cos nx + b_n \sin nx). \\ &= \frac{\sinh \pi}\pi \left( 1 + 2\sum_{n=1}^\infty (-1)^n \frac {\cos nx} {n^2 + 1}\right) \end{align*} \]

Substituting \(x = \pi\) then gives:

\[ \cosh \pi = \frac{\sinh \pi}\pi \left( 1 + 2\sum_{n=1}^\infty \frac 1 {n^2 + 1}\right) \]

which gives Bob's answer. – C Lim · 4 years ago

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Here's a shorter way.

Note that \( \cot( x)=\frac{1}{x}+ \sum_{k=1}^{\infty} \frac{2x}{x^2-k^2 \pi^2} \) (Proof: use infinite product for sine, take natural logs, differentiate, rearrange). Thus,

\(\large \frac{1}{\pi^2} \sum_{k=1}^{\infty} \frac{2x}{\frac{x^2}{\pi^2}-k^2 }=\cot( x)-\frac{1}{x}\)

\(\large \sum_{k=1}^{\infty} \frac{x}{k^2-\frac{x^2}{\pi^2} }=\frac{-\pi^2}{2} \left(\cot( x)-\frac{1}{x}\right)\)

Let \(x=i\pi\),

\(\large \sum_{k=1}^{\infty} \frac{i \pi}{k^2+1 }=\frac{-\pi^2}{2} \left(\cot( i \pi)-\frac{1}{i \pi}\right)\)

\(\large \sum_{k=1}^{\infty} \frac{1}{k^2+1 }=\frac{i\pi}{2} \left(\cot( i \pi)-\frac{1}{i \pi}\right)\)

\(\large \sum_{k=1}^{\infty} \frac{1}{k^2+1 }=\frac{i\pi}{2} \frac{\cos(i \pi)}{\sin(i \pi)}-\frac{1}{2}\)

\(\large \sum_{k=1}^{\infty} \frac{1}{k^2+1 }=\frac{i\pi}{2} \frac{\cosh( \pi)}{i\sinh( \pi)}-\frac{1}{2}\)

\(\large \sum_{k=1}^{\infty} \frac{1}{k^2+1 }=\frac{\pi}{2} \frac{\cosh( \pi)}{\sinh( \pi)}-\frac{1}{2}\) – A L · 4 years ago

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Mathematica gave \(\frac{-1}{2} + \frac{\pi}{2}\coth \pi\). I know this isn't exactly the point, but sometimes the answer is the best place to start a question like this. – Bob Krueger · 4 years ago

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Is there a systematic way to approach these questions? I've seen Fourier Series solutions before, but they always struck me as most likely discovered by accident. For a simpler example, why would you immediately think to consider the Fourier Series of \(f(x) = |x|\) if you wanted to find \(\zeta(2)\)? – Eric Edwards · 4 years ago

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– A L · 4 years ago

The second answer here http://math.stackexchange.com/questions/8337/different-methods-to-compute-sum-limits-n-1-infty-frac1n2 makes the choice much more obvious (albeit it uses the function \(x^2\), not \(|x|\), although of course these choices would yield identical results, just square at the end).Log in to reply

Excellent work, everybody! – Ahaan Rungta · 4 years ago

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