Does the series $\sum_{n=1}^{\infty}\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+x}}$ represent a well-known function? Consider the function series 
$$f(x)=\sum_{n=1}^{\infty}\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+x}}$$
According to some theorems, I found that the above series is convergent point wise on $(-1, +\infty)$. 
Does the series represent a well-known function?
 A: While the comments already offer an answer to the question as stated, the question could be understood in a more broad context (as I'm sure the OP would agree): What do we know (or can find out) about this function?
$$f(x)=\sum_{n=1}^{\infty} \left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+x}} \right)$$
Turns out, it's quite a lot.
First, let us point out a very useful property, which is immediately seen from the series definition:

$$f(x+1)=f(x)+\frac{1}{\sqrt{x+1}} \tag{1}$$

Which conveniently means that we only need to define the function for $x \in (-1,1)$ and then use (1) to extend it to the whole domain $(-1, + \infty)$.
Now let us point out the trivial value:
$$f(0)=0$$
We will need it, because it is easier to consider the derivative of this function. The fact that $f'(x)$ can be found by differentiating the series by term is left without proof here, but it is confirmed by Mathematica (exact evaluation in terms of Hurwitz zeta function).
The derivative is represented by a convergent series:
$$f'(x)=\frac{1}{2} \sum_{n=1}^{\infty} \frac{1}{(n+x)^{3/2}}$$
Using the fact that $|x|<1$ we rewrite the series:
$$f'(x)=\frac{1}{2} \sum_{n=1}^{\infty}\frac{1}{n^{3/2}} \frac{1}{(1+x/n)^{3/2}}=\frac{1}{2} \sum_{n=1}^{\infty} \frac{1}{n^{3/2}} \sum_{k=0}^{\infty} \binom{-3/2}{k} \frac{x^k}{n^k}$$
Changing the order of summation:
$$f'(x)=\frac{1}{2} \sum_{k=0}^{\infty} \binom{-3/2}{k} x^k \sum_{n=1}^{\infty} \frac{1}{n^{k+3/2}}=\frac{1}{2} \sum_{k=0}^{\infty} \binom{-3/2}{k} \zeta \left(k+\frac{3}{2} \right) x^k$$
We have obtained the Taylor series for the derivative. Integrating them by term and using the known value for $f(0)$ we have:
$$f(x)=\frac{x}{2} \sum_{k=0}^{\infty} \binom{-3/2}{k} \zeta \left(k+\frac{3}{2} \right) \frac{x^k}{k+1}$$
Rewriting the binomial coefficient using the Gamma function identities, gives us:

$$f(x)=\frac{x}{2} \sum_{k=0}^{\infty} (-1)^k \frac{(2k+1)!}{k! ~(k+1)!} \zeta \left(k+\frac{3}{2} \right) \frac{x^k}{4^k} \tag{2}$$

This is Taylor series which defines the function for $x \in (-1,1)$, and together with $(1)$ for the rest of its domain.
This is the plot for the original series (red) and for the series (2) (gray) which show perfect agreement with each other. Also, we can see that the function is quite close to $x$ for small arguments.


Using the integral definition of the zeta function:
$$\zeta(s)=\frac{1}{\Gamma(s)} \int_0^\infty \frac{t^{s-1} dt}{e^t-1}$$
We obtain another defining expression for the series:

$$f(x)=\frac{2}{\sqrt{ \pi}} \int_0^\infty \frac{1-e^{-x t^2}}{e^{t^2}-1}dt \tag{3}$$

Which converges for all $x \in (-1, \infty)$.

An obvious generalization is:
$$f_r(x)=\sum_{n=1}^{\infty} \left(\frac{1}{n^r}-\frac{1}{(n+x)^r} \right), \qquad r>0$$
Which, in the same way can be written as:
$$f_r(x)=r x \sum_{k=0}^{\infty} \binom{-r-1}{k} \zeta \left(k+1+r \right) \frac{x^k}{k+1}$$
$$f_r(x+1)=f_r(x)+\frac{1}{(n+x)^r}$$
