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Assuming the series representation of the Bessel function $J_v(x)$

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I was given that the radius of convergence was ∞. I've tried using the ratio test but I dunno how the gamma function would disappear...Would someone be able to explain it to me?

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    $\begingroup$ $\Gamma\left((m+1)+\nu+1\right) = (m+\nu+1)\Gamma(m+\nu+1)$. $\endgroup$ May 11 '14 at 11:33
  • $\begingroup$ how does that disappear in the ratio test? $\endgroup$
    – Kennan
    May 11 '14 at 11:50
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    $\begingroup$ Don't worry about the gamma function. It is very big in itself, which contributes to the convergence of the series, and moreover it is surrounded by big things: factorials, exponentials. You can easily get rid of the Gamma function with a lower bound. Even the incredibly crude one $$\Gamma (m+\nu+1)\ge 1$$ is enough here. $\endgroup$ May 11 '14 at 12:05
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The ratio of (the modulus of) two consecutive terms is

$$\frac{\lvert x\rvert^{2(m+1)}}{2^{2(m+1)+\nu} (m+1)!\Gamma\left((m+1)+\nu+1\right)}\frac{2^{2m+\nu}m!\Gamma(m+\nu+1)}{\lvert x\rvert^{2m}} = \frac{\lvert x\rvert^2}{4(m+1)}\frac{\Gamma(m+\nu+1)}{\Gamma\left((m+1)+\nu+1\right)}$$

and with the functional equation $\Gamma(z+1) = z\Gamma(z)$, this reduces to

$$\frac{\lvert x\rvert^2}{4(m+1)(m+\nu+1)},$$

which is easily seen to converge to $0$ for every $x$.

(Note: if $\nu$ is a negative integer, the terms of the series for $m < -\nu$ are all $0$, so the ratio test shall be applied only for $m \geqslant -\nu$ then.)

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