Is there a closed form sum of 


$\sum\_{k=0}^{\infty} \frac{x^k}{(k!)^2}$


It is trivial to show that it is less than $e^x$ but is there a tighter bound?


Thanks



Summing the infinite series $\sum_{k=0}^{\infty} \frac{x^k}{(k!)^2}$
Here there are many possibilities. One of them is as follows. Note that for $k=0,1,\dots$
\begin{equation}
 \frac1{(k!)^2}=\binom{2k}k\,\frac1{(2k)!}\le\frac{2^{2k}}{(2k)!}, 
\end{equation}
whence for $x\ge0$ the sum of your series is no greater than 
\begin{equation}
 B(x):=\sum\_{k=0}^{\infty} \frac{(4x)^k}{(2k)!}=\cosh\sqrt{4x},
\end{equation}
which is much less than $e^x$ for large $x$. 


**Added:** As pointed out in the comment by Carlo Beenaker, 
\begin{equation}
 S(x):=\sum\_{k=0}^{\infty} \frac{x^k}{(k!)^2}\sim e^{\sqrt{4x}}/(4\pi\sqrt x)^{1/2} 
\end{equation}
and hence 
\begin{equation}
 \ln B(x)\sim\ln S(x)
\end{equation}
as $x\to\infty$; that is, the bound $B(x)$ on $S(x)$ is logarithmically asymptotically tight for large $x$ (in contrast with the bound $e^x$). 


