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There are infinitely many real solutions to the equation $\tan x=x$. Denote the increasing sequence of positive solutions by $ (\lambda_n)_{n=1}^{\infty}$. I want to evaluate the sum

$$ \sum_{n=1}^{\infty} \frac{1}{\lambda_n^2} \hspace{1cm} (1) $$ (the series can be seen to converge by using a crude approximation such as $\lambda_n \approx \frac{\pi(2n+1)}{2}$).

To find the sum I'm thinking to use Hadamard's product theorem applied to a function whose roots are $\lambda_n$ and then to expand the product and compare coefficients as in Euler's evaluation of $\zeta(2)$. The Hadamard product theorem states that any entire function $f(z)$ of growth order $\alpha<2$ can be written as $$ f(z) = z^m e^{Az+B} \prod_{\rho} e^{z/\rho} \left(1-\frac{z}{\rho} \right) $$ where the product is over all the non-zero complex roots $\rho$ of $f$ repeated according to multiplicity, $m \geq 0$ is the order of the root of $f$ at $z=0$, and $A,B$ are two constants to be determined.

Now $ \tan z-z $ is not entire but it has the same non-zero roots as the function $ f(z) = \sin z - z \cos z$ which is entire. These are all real roots of order 1 and they are symmetric since the function is odd. Thus the non-zero roots of $f(z)$ are at $\pm \lambda_n$ and there is also a triple root at $z=0$. Further, the function has growth order 1, so applying Hadamard's product formula gives $$ \sin z-z \cos z = e^{Az+B} z^3 \prod_{n=1}^{\infty} e^{z/\lambda_n} \left(1-\frac{z}{\lambda_n} \right) e^{z/(-\lambda_n)} \left(1-\frac{z}{(-\lambda_n)} \right) = e^{Az+B} z^3 \prod_{n=1}^{\infty} \left(1-\frac{z^2}{\lambda_n^2} \right) $$ and we may compute $$e^B = \lim_{z \to 0} \frac{\sin z-z\cos z}{z^3} = \frac{1}{3} $$ and $A=0$ since $\frac{\sin z-z\cos z}{z^3}$ is an even function. Therefore $$ \sin z-z \cos z = \frac{z^3}{3} \prod_{n=1}^{\infty} \left(1-\frac{z^2}{\lambda_n^2} \right) \hspace{1cm} (2) $$ We may now expand the left side in a Taylor series and compare coefficients with the right side to obtain $$ \sum_{n=1}^{\infty} \frac{1}{\lambda_n^2} = \frac{1}{10} $$ as well as higher order sums such as $$ \sum_{n=1}^{\infty} \frac{1}{\lambda_n^4} = \frac{1}{350} $$ and so on. All sums of reciprocals of even powers will be rational numbers.

My questions are:

  1. Is this solution correct?

  2. Are there other ways of evaluating the series $(1)$ without using the Hadamard product?

  3. Are there any general techniques (contour integration for example) available for evaluating sums of the form $$ \sum_{\rho} g(\rho) $$ where the sum is over all roots $\rho$ of an entire function $f$ and $g$ is some other analytic function for which the series converges? Thanks!

Dave
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    The same value $\sum_{n=1}^{\infty} \frac{1}{\lambda_n^2} = \frac{1}{10}$ is also found in several answers here https://math.stackexchange.com/q/75206/42969, so that should be correct. – Martin R Dec 30 '23 at 12:14
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    I think that the product formula (2) cannot give $\sum \lambda_n^{-4}$ but $\sum_{n\ne m} \lambda_n^{-2} \lambda_m^{-2}$. Considering $\lambda_n$ as the zeros of a Bessel function: $J_{3/2}(\lambda_n)=0$, expressions exist for the sum of their inverse even powers as discussed here (a recent analysis can be found in a paper by Jorge L. deLyra). One obtains $\sum \lambda_n^{-4}=1/350$. – Paul Enta Dec 30 '23 at 13:51
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    @PaulEnta I expanded the product and equated coefficients of $z^7$ in (2) which gives $\sum_{n>m} \frac{1}{\lambda_n^2\lambda_m^2}$. Comparing with $\left(\sum_{n=1}^{\infty} \frac{1}{\lambda_n^2}\right)^2$ we can rearrange for the diagonal of the double sum, which is the sum of fourth powers. Maybe I made a computation error. – Dave Jan 02 '24 at 09:10
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    Sorry @Dave, I was wrong! Actually, the decomposition you mention - the coefficient of $z^7$ is $1/3\times1/2\left( \left( \sum\lambda_n^2 \right)^2-\sum\lambda_n^4 \right)$ - gives $1/350$ as expected. – Paul Enta Jan 02 '24 at 13:09
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    @PaulEnta Thanks! I corrected it – Dave Jan 02 '24 at 22:48

1 Answers1

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Too long for a comment.

We can have interesting results for $$S_{2p}=\sum_{n=1}^{\infty} \frac{1}{\lambda_n^{2p}}$$ using as a better approximation $$\lambda_n= (2 n+1)\frac{\pi}{2} -\frac{2}{\pi (2 n+1)}$$ They write $$S_{2p}=-\left(\frac{2 \pi }{\pi ^2-4}\right)^{2p}+\frac 18\,\sec ^{2 p}(1)\,\, T_{2p}$$ with $$T_2=2+\sin (2) $$ $$T_4=\frac{44+42 \sin (2)-3 \sin (4)-4 \cos (2)}{96} $$ $$T_6=\frac{3732+4545 \sin (2)-540 \sin (4)+45 \sin (6)-944 \cos (2)-836 \cos (4)}{30720}$$ and so on.

For the odd powers, the results express in terms of polygamma functions such as $$S_3=\frac{3}{16 \pi ^2}\left(\psi ^{(1)}\left(\frac{3}{2}-\frac{1}{\pi }\right)-\psi ^{(1)}\left(\frac{3}{2}+\frac{1}{\pi }\right)\right)-$$ $$\frac{1}{16 \pi ^3}\left(\psi ^{(2)}\left(\frac{3}{2}-\frac{1}{\pi }\right)+\psi ^{(2)}\left(\frac{3}{2}+\frac{1}{\pi }\right)\right)$$