What is $\sum_{k=0}^\infty {k^k \over (k!)^2}$? I know that this series converges, and the limit is approximately 3.548.
But what is the exact sum, and how do you determine it?
 A: There is no clear answer that would not involve some by itself sophisticated function.
However, there is one curious thing about it that may reveal some properties and at least the error term regarding the summation.
Let us write the sum as $$\sum\limits_{k=0}^{\infty}\frac{1}{k!}\frac{k^k}{k!} \;  (1)$$
This is indeed like Taylor of some strange function that has evaluation of derivatives together with standard term $(x-x_{0})^n$ as $\frac{k^k}{k!}$. Let us try to find something similar to it.
Notice the very important property about two successive terms
$$ \frac{\frac{(k+1)^{k+1}}{(k+1)!}}{\frac{k^k}{k!}}=(1+\frac{1}{k})^k$$ and these are surprise surprise! the famous series for $e$ itself.
This helps in many ways. First you can check the convergence of the series itself since 
$$\sum\limits_{k=0}^{\infty}\frac{e^k}{k!}=e^e$$
and we know that our series is lower than this one since $(1+\frac{1}{k})^k$ is reaching $e$ from below.
Next we have a fast test over the size of the error term $\sum\limits_{k=m}^{\infty}\frac{e^k}{k!}$ seeing that we need first hardly 50 terms to get a high precision result.
You can continue in various directions getting more information about the series and where it came from. For me it would be very interesting to understand the form of the function involved in the Taylor series. For example, if you write
$$f(e)=f(0)+\sum_{k=1}^{\infty}\frac{1}{k!}\frac{k^k}{e^k k!}e^k$$
suddenly the problem turns into finding a function that has derivatives at 0 as $f^{(k)}(0)=\frac{k^k}{e^k k!}$
Using various approximations you can reach some sort of better orientation of the final value and which factor is reducing the direct estimation of $e^e$ which is the only one we have got in a precise manner so far.
