Power series $\sum\limits_{n=0}^\infty \frac{x^n}{(n!)^2}$

Question

It is well known that $\sum\limits_{n=0}^\infty \frac{x^n}{(n!)} = e^x$.

What is $\sum\limits_{n=0}^\infty \frac{x^n}{(n!)^2}$ ?

I have no idea about this problem. This problem raises when I calculate an integral of the Bessel function. — Muses_China, Nov 27 '20 at 12:07
Maybe it could be represented as some transcendental functions... — Muses_China, Nov 27 '20 at 12:10
Does this answer your question? Closed form of series with factorial-squared denominator? – found with Approach0 — Martin R, Nov 27 '20 at 12:15

score 1 · Accepted Answer · answered Nov 27 '20 at 12:13

1

$$\sum\limits_{n=0}^\infty \frac{x^n}{(n!)^2}=I_0\left(2 \sqrt{x}\right)$$ where appears the modified Bessel function of the first kind.

answered Nov 27 '20 at 12:13

Claude Leibovici

Great thanks! I am a freshman to Bessel function, this qustion may seem silly to you! – Muses_China Nov 27 '20 at 12:16
@Muse_China. There is no silly question but there are silly answers ! Welcome to the fascinating world of Bessel functions ! – Claude Leibovici Nov 27 '20 at 12:18

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