What happens when I convert a Taylor series into an integral? Suppose we have the Taylor series of an analytic function as follows:
$$f(x) = \sum_{k=0}^\infty \frac{1}{k!} a_k x^k$$
Then I decide to (kind of) turn it into an integral:
$$g(x) = \int_0^\infty \frac{1}{\Gamma(k+1)} a(k) x^k \, dk$$
Clearly, $f(x) \neq g(x)$.  But the values the two functions produce are somewhat close to each other.  What's the relation between the two?
 A: I'm assuming that you define $a(k)=a_{[k]}$, where $[k]$ is the floor function. You could similarly define it with a ceiling function and the answer should remain the same. First review the statement and proof of the integral test. 
The gamma function is strictly increasing in $k$, and lets assume $a_kx^k/k!$ is decreasing (it has to go to zero regardless), so in this case we get something like:
$$f(x)-a_0=\sum_{k\geq 1}\frac{a_k}{k!}x^k\leq \int_0^\infty \frac{a(k)}{\Gamma(k+1)}x^kdk \leq \sum_{k\geq 0} \frac{a_k}{k!}x^k=f(x).$$
Note that if the terms are not monotonically decreasing then you'll get considerably more deviation. 
Be care with the $a_0$ term. If $a_0=0$ for example, then by our monotonicity assumption this would mean all other terms are zero. So if you want to correct for this, you need to further offset the sum to do the integral/sum comparision.
A: I believe that the summation is an approximation for the integral. Basically, the summation only counts up in whole number increments. The integral counts all values between these numbers all the way up to infinity. This means that the sum will approximate the integral. 
Here's a visual representation: 

The sum is the summation of the areas of all the green rectangle. The integral is the area under the curve. That's why the two values are similar but not the same. 
Image source: Integration and the Fundamental Theorem of Calculus by 3blue1brown
A: I think you can use the Abel's summation formula  to get the equality between the two functions as it allows us to transform a series to integral.
