I was investigating methods for series acceleration and I found this identity: $$e=\sum _{n=0}^{\infty } \left(\sqrt{\frac{\pi }{2}} (2 n+1)\right) I_{n+\frac{1}{2}}(1)$$ where $I$ is the modified Bessel function of the first kind.
Could you explain me where this identity comes from ?
I tried computing the Fourier-Bessel series for the exponential function but it did not work.
Since: $$I_{1/2}(z)=\sqrt{\frac{2}{\pi z}}\,\sinh z$$ and by the multiplication formula:
$$\lambda^{-\nu} I_\nu (\lambda z) = \sum_{n=0}^\infty \frac{1}{n!} \left(\frac{(\lambda^2-1)z}{2}\right)^n I_{\nu+n}(z)\tag{1} $$ we have:
$$\sqrt{\frac{2}{\pi}}\,\frac{\sinh \lambda}{\lambda} = \sum_{n=0}^\infty \frac{1}{n!} \left(\frac{\lambda^2-1}{2}\right)^n I_{n+1/2}(1),\tag{2} $$
$$\sqrt{\frac{2}{\pi}}\,\frac{\sinh \cosh u}{\cosh u} = \sum_{n=0}^\infty \frac{\sinh^{2n}(u)}{2^n\cdot n!} I_{n+1/2}(1),\tag{3} $$
$$\sqrt{\frac{2}{\pi}}\,\frac{\sinh\sqrt{1+2t^2}}{\sqrt{1+2t^2}} = \sum_{n=0}^\infty \frac{t^{2n}}{n!} I_{n+1/2}(1),\tag{4} $$
$$\sqrt{\frac{2}{\pi}}\,\frac{\sinh\sqrt{1+2w}}{\sqrt{1+2w}} = \sum_{n=0}^\infty \frac{w^{n}}{n!} I_{n+1/2}(1).\tag{5} $$ Now replacing $w$ with $\mu\eta$ and interating both sided against $e^{-\mu}$ over $\mathbb{R}^+$ we get: $$\int_{0}^{+\infty}\frac{\sinh\sqrt{1+2\mu\eta}}{\sqrt{1+2\mu\eta}}e^{-\mu}\,d\mu = \sqrt{\frac{\pi}{2}}\sum_{n=0}^\infty \eta^n I_{n+1/2}(1),\tag{6} $$ $$\int_{0}^{+\infty}\frac{\sinh\sqrt{1+2\mu\eta^2}}{\sqrt{1+2\mu\eta^2}}\eta e^{-\mu}\,d\mu = \sqrt{\frac{\pi}{2}}\sum_{n=0}^\infty \eta^{2n+1} I_{n+1/2}(1),\tag{7} $$ $$\sum_{n=0}^\infty (2n+1)\, I_{n+1/2}(1)=\sqrt{\frac{2}{\pi}}\frac{d}{d\eta}\left.\int_{0}^{+\infty}\frac{\sinh\sqrt{1+2\mu\eta^2}}{\sqrt{1+2\mu\eta^2}}\eta e^{-\mu}\,d\mu\right|_{\eta=1}\tag{8}$$
$$ \sum_{n=0}^\infty (2n+1)\, I_{n+1/2}(1)=\sqrt{\frac{2}{\pi}}\int_{0}^{+\infty}\frac{2m\sqrt{1+2m}\cosh\sqrt{1+2m}+\sinh\sqrt{1+2m}}{(2m+1)^{3/2}}e^{-m}\,dm\tag{9}$$ and the last integral is not so difficult to compute. It is: $$ I=\int_{1}^{+\infty}\frac{e^{\frac{1}{2}-\frac{u^2}{2}} \left(u \left(-1+u^2\right) \cosh u+\sinh u\right)}{u^2}\,du$$ or, integrating by parts: $$ I=\cosh 1-\int_{1}^{+\infty}\left(u\cosh u+(1-u^2)\sinh u\right)e^{\frac{1-u^2}{2}}\,du\tag{10} $$ $$ I = \cosh 1+\frac{e}{2}\int_{0}^{+\infty}(v^2+v-1)\,e^{-\frac{v^2}{2}}\,dv-\frac{1}{2e}\int_{0}^{+\infty}(v^2+3v+1)e^{-\frac{v(v+4)}{2}}\,dv\tag{11}$$ so: $$ I = \cosh 1+\frac{e}{2}-\frac{1}{2e} = \color{red}{e} \tag{12}$$ giving, due to $(9)$:
$$ \sqrt{\frac{\pi}{2}}\sum_{n=0}^{+\infty}(2n+1)\, I_{n+1/2}(1) = e\tag{13}$$
as wanted.
Hi Ill give you another series for E more faster $$e=\frac{2 \sum _{n=1}^{\infty } \pi I_n(1)}{\pi }+I_0(1)$$ but i could give you a demostration at the moment happy new year
good work Jack but the Integral it is very difficult to resolve anyway i try to get the series of the Sinh(x)/x using series Bessel Fourier and we get a different seriesthan you give it above i think the following is it faster
$$\frac{\sinh (x)}{x}=\sum _{n=0}^{\infty } I_{n+\frac{1}{2}}(1) \left(\pi 2^{n-\frac{7}{2}} \left((-1)^n+1\right) (2 n+1) x^n \left(\frac{(-1)^{\left\lfloor \frac{n}{2}\right\rfloor +1} 4^{-\left\lfloor \frac{n}{2}\right\rfloor } x^{-2 \left(\left\lfloor \frac{n}{2}\right\rfloor +1\right)} \sec \left(\pi \left(n-\left\lfloor \frac{n}{2}\right\rfloor \right)\right) \, _3\tilde{F}_2\left(1,-\frac{n}{2}+\left\lfloor \frac{n}{2}\right\rfloor +\frac{1}{2},-\frac{n}{2}+\left\lfloor \frac{n}{2}\right\rfloor +1;\left\lfloor \frac{n}{2}\right\rfloor +2,-n+\left\lfloor \frac{n}{2}\right\rfloor +\frac{3}{2};\frac{1}{x^2}\right)}{\Gamma \left(n-2 \left\lfloor \frac{n}{2}\right\rfloor \right)}+\frac{4 \sec (\pi n) \, _2\tilde{F}_1\left(-\frac{n}{2}-\frac{1}{2},-\frac{n}{2};\frac{1}{2}-n;\frac{1}{x^2}\right)}{\Gamma (n+2)}\right)\right)$$ but it is very complicate thanks anyway.