Proof of Some Known Formulas

Up to this point $p$ could be any strictly positive real number. From now on, however, we have to assume that $p$ is a strictly positive integer. In order to further simplify the expression obtained above, in general it will be necessary to write $J_{\nu+p}(\xi_{\nu k})$ in terms of $J_{\nu+1}(\xi_{\nu k})$, which can be done using the recurrence formula of the Bessel functions [6], so long as $p$ is an integer. Let us examine a few of the initial cases. For $p=1$ we have simply

$$\frac{\Gamma(\nu+1)}{2^{2}\Gamma(\nu+2)} = \sum_{k=1}^{\infty} \frac{1}{\xi_{\nu k}^{2}},$$ (36)

so that the formula for the sum that corresponds to this case is

$$\sigma(1,\nu) = \sum_{k=1}^{\infty} \frac{1}{\xi_{\nu k}^{2}}$$

$$= \frac{1}{2^{2}(\nu+1)},$$ (37)

were we used the properties of the gamma function, thus obtaining a polynomial on $\nu$ in the denominator. In this way we obtain the first of the known results, and this formula is therefore proven, being valid for any non-negative real value of $\nu$. For $p=2$ we have

$$\frac{\Gamma(\nu+1)}{2^{3}\Gamma(\nu+3)} = \sum_{k=1}^{\i... ...{3}}\, \frac{J_{\nu+2}(\xi_{\nu k})}{J_{\nu+1}(\xi_{\nu k})}.$$ (38)

In order to simplify this expression we write the recurrence formula as

$$J_{\nu+2}(\xi) = \frac{2(\nu+1)}{\xi}\,J_{\nu+1}(\xi)-J_{\nu}(\xi),$$ (39)

where we exchanged $\nu$ for $\nu+1$. Applying this for $\xi=\xi_{\nu k}$ and using once more the fact that $J_{\nu}(\xi_{\nu k})=0$, we get

$$J_{\nu+2}(\xi_{\nu k}) = \frac{2(\nu+1)}{\xi_{\nu k}}\,J_{\nu+1}(\xi_{\nu k}),$$ (40)

so that in this case we have

$$\frac{\Gamma(\nu+1)}{2^{3}\Gamma(\nu+3)} = 2(\nu+1) \sum_{k=1}^{\infty} \frac{1}{\xi_{\nu k}^{4}},$$ (41)

from which it follows that the formula for the sum that corresponds to this case is

$$\sigma(2,\nu) = \sum_{k=1}^{\infty} \frac{1}{\xi_{\nu k}^{4}}$$

$$= \frac{1}{2^{4}(\nu+1)^{2}(\nu+2)}.$$ (42)

We thus obtain the second known result, which is now proven, and which also has a polynomial on $\nu$ in the denominator. The proof of the first two formulas is therefore quite straightforward. In the $p=3$ case, however, something slightly different happens. In this case we have

$$\frac{\Gamma(\nu+1)}{2^{4}\Gamma(\nu+4)} = \sum_{k=1}^{\i... ...{4}}\, \frac{J_{\nu+3}(\xi_{\nu k})}{J_{\nu+1}(\xi_{\nu k})},$$ (43)

and hence we must write one more version of the recurrence formula. Exchanging $\nu$ for $\nu+2$ in the original formula, and applying at $\xi=\xi_{\nu k}$, we obtain

$$J_{\nu+3}(\xi_{\nu k}) = \frac {2(\nu+2)} {\xi_{\nu k}}\,J_{\nu+2}(\xi_{\nu k})-J_{\nu+1}(\xi_{\nu k}).$$ (44)

Substituting in this the solution found for the previous case, which gives us $J_{\nu+2}(\xi_{\nu k})$ in terms of $J_{\nu+1}(\xi_{\nu k})$, we get

$$J_{\nu+3}(\xi_{\nu k}) = \left[ \frac{2^{2}(\nu+1)(\nu+2)}{\xi_{\nu k}^{2}} - 1 \right] J_{\nu+1}(\xi_{\nu k}).$$ (45)

In this way we get in this case the result

$$\frac{\Gamma(\nu+1)}{2^{4}\Gamma(\nu+4)} = \sum_{k=1}^{\infty} \frac{1}{\xi_{\nu k}^{4}} \left[ \frac{2^{2}(\nu+1)(\nu+2)}{\xi_{\nu k}^{2}} - 1 \right]$$

$$= 2^{2}(\nu+1)(\nu+2) \sigma(3,\nu) - \sigma(2,\nu).$$ (46)

We see that in this case a linear combination of the sums of two different powers of the zeros $\xi_{\nu k}$ appears. Using the properties of the gamma function and substituting the value obtained previously for $\sigma(2,\nu)$, we get for $\sigma(3,\nu)$

$$\sigma(3,\nu) = \frac{2(\nu+2)}{2^{6}(\nu+1)^{3}(\nu+2)^{2}(\nu+3)}.$$ (47)

The formula for the sum that corresponds to this case is therefore

$$\sigma(3,\nu) = \sum_{k=1}^{\infty} \frac{1}{\xi_{\nu k}^{6}}$$

$$= \frac{1}{2^{5}(\nu+1)^{3}(\nu+2)(\nu+3)},$$ (48)

where we once more have a polynomial on $\nu$ in the denominator. We thus obtain the third of the known results, and the formula for $\sigma(3,\nu)$ is proven.

It is clear that we can proceed in this way indefinitely, thus obtaining the formulas for successive values of $p$. In each case it is necessary to first use the recurrence formula in order to write $J_{\nu+p}(\xi)$ in terms of $J_{\nu+1}(\xi)$. In general the result will be a linear combination of sums of several distinct inverse powers of $\xi_{\nu k}$. At this point the use of the general formula in Eq. (35) will produce an expression for the linear combination of the corresponding sums $\sigma(p,\nu)$. Finally, it is necessary to solve the resulting expression for the sum with the largest value of $p$ so far, using for this end the results obtained previously for the other sums. In this way all the formulas for the sums $\sigma(p,\nu)$ can be derived successively by purely algebraic means, resulting every time in the ratio of two polynomials, with the one in denominator completely factored.