I don't think there's an analytical solution, but you can approximate it.
First approximation is integral,
Next approximations would involve some play with operators.
Let's define D\equiv \frac{d}{dx} as the derivative operator, and observe the operator \sigma \equiv e^{D}
Functions of operators are defined via their Taylor Series, so:
e^{D} f=\sum^{\infty}_{k=0}\frac{D^{k}f(x)}{k!}=\sum^{\infty}_{k=0}\frac{1}{k!}\frac{d^{k}f(x)}{dx^{k}}(x+1-x)^k=f(x+1)
(Notice that the last sum is the Taylor Series of f(z) around the point x)
So \sigma is some sort of a delay\shift operator.
Let's now observe the operator S defined as
F(n)=Sf=\sum^{n}_{k=1}f(k)
You can notice that \sigma Sf=\sum^{n+1}_{k=1}f(k)
So \sigma S f-Sf=f(n+1)=\sigma f
And conclude S(\sigma -1)=\sigma
And if you recall the definition of \sigma then:
S=(1-\sigma^{-1})^{-1}=\frac{1}{1-e^{-D}}
S(D-\frac{D^{2}}{2}+\frac{D^{3}}{6}-...)=1
Knowing the expansion of the exponential function, you can approximate the Taylor expansion of S. The interesting fact is that S has a simple pole at "D=0", which means that in the expansion of S you will have a negative power of D:
S=D^{-1}+...
Which means that integration (the inverse of D) is the first approximation to discrete summation.
A better approximation would be
S=D^{-1}+\frac{1}{2}+\frac{1}{4}D
Operate S on \frac{1}{\sqrt{n}}
And get:
\sum^{n}_{k=2}\frac{1}{\sqrt{k}}=\int^{n}_{k=2}\frac{dk}{\sqrt{k}}+\frac{1}{2\sqrt{n}}-\frac{1}{8n\sqrt{n}}
And after some tiresome calculation for n=361
=35.1977
Which I think is a pretty good approximation, if only to show that the operator thing is correct.
Taking further derivatives would make a better approximation.
