Simplifying a Summation : $\sum_{i=0}^\infty \frac{4^i}{(i!)^2}$

Question

How can I simplify the following summation:

$$\sum_{i=0}^\infty \frac{4^i}{(i!)^2}$$

I cannot think of anyway to simplify this.

Answer 1

As said in comments, since the expansion of the modified Bessel function of the first kind write (see here)$$I_\nu(z)=\left(\frac z2 \right)^\nu \,\sum_{k=0}^\infty \frac{\left(\frac z2 \right)^{2k} }{k! \,\Gamma(\nu+k+1)}$$ then $$I_o(z)=\sum_{k=0}^\infty \frac{\left(\frac z2 \right)^{2k} }{k! \,\Gamma(k+1)}=\sum_{k=0}^\infty \frac{\left(\frac z2 \right)^{2k} }{(k!)^2 }$$ and, as a result, $$\sum_{i=0}^\infty \frac{x^i}{(i!)^2}=I_0\left(2 \sqrt{x}\right)$$ which cannot reduce to elementary functions.

For your case, the numerical evaluation would converge very fast since $$u_n=\frac {4^n}{(n!)^2}\implies \frac{u_{n+1}}{u_n}=\frac{4}{(n+1)^2}$$ I give you below the numerical values for suucessive $n$ $$\left( \begin{array}{ccc} n & \text{exact} & \text{approx} \\ 0 & 1 & 1 \\ 1 & 5 & 5 \\ 2 & 9 & 9 \\ 3 & \frac{97}{9} & 10.7777777777778 \\ 4 & \frac{101}{9} & 11.2222222222222 \\ 5 & \frac{847}{75} & 11.2933333333333 \\ 6 & \frac{4577}{405} & 11.3012345679012 \\ 7 & \frac{1121429}{99225} & 11.3018795666415 \\ 8 & \frac{373811}{33075} & 11.3019198790627 \\ 9 & \frac{90836089}{8037225} & 11.3019218697996 \\ 10 & \frac{2270902241}{200930625} & 11.3019219494291 \\ 11 & \frac{1221240761}{108056025} & 11.3019219520614 \\ 12 & \frac{353287505863}{31259064375} & 11.3019219521346 \\ 13 & \frac{417939119435993}{36979473155625} & 11.3019219521363 \end{array} \right)$$ As you can see, using seven terms, we already obtain six correct significant figures.