Does the sum of the series $\sum_{n=1}^{\infty}\frac{\sqrt{n+1}-\sqrt n}{n}$ have an analytic expression? Just out of curiosity, I'd like to know whether or not the sum of the series
$$\sum_{n=1}^\infty \frac{\sqrt{n+1}-\sqrt n}{n}$$
has a known analytic expression.
I stumbled across this series while trying to evaluate
$$\int_1^\infty\frac{1}{\lfloor x^2\rfloor}dx$$
The convergence of this integral can be seen by making use of the inequality $\lfloor x\rfloor > x-1$ and the fact that $\coth^{-1}(t)\to 0$ as $t\to\infty$:
\begin{align*}
\int_1^t\frac{1}{\lfloor x^2\rfloor}dx &= \int_\sqrt{1}^\sqrt{2}\frac{1}{\lfloor x^2\rfloor}dx+\int_\sqrt{2}^t\frac{1}{\lfloor x^2\rfloor}dx\\
&< \int_\sqrt{1}^\sqrt{2}\frac{1}{\lfloor x^2\rfloor}dx+\int_\sqrt{2}^t\frac{1}{x^2-1}dx\\
&= \int_\sqrt{1}^\sqrt{2}\frac{1}{1}dx-\int_\sqrt{2}^t\frac{1}{1-x^2}dx\\
&= \sqrt{2}-1-\left[\coth^{-1}(t)-\coth^{-1}\left(\sqrt 2\right)\right]\\
&= \sqrt{2}-1-\coth^{-1}(t)+\coth^{-1}\left(\sqrt 2\right)\\
&\to \sqrt{2}-1+\coth^{-1}\left(\sqrt 2\right)\text{ as }t\to\infty\\
\end{align*}
Since this implies that $\int_1^t 1/\lfloor x^2\rfloor dx$ is strictly increasing ($1/\lfloor x^2\rfloor >0$ for every $x\geq 1$) and bounded above, the integral necessarily converges. By breaking up the integral
$$\int_1^\sqrt{k+1}\frac{1}{\lfloor x^2\rfloor}dx$$
into integrals indexed by the intervals $\left[\sqrt{i},\sqrt{i+1}\right]$ for $i=1,2,3,...,k$ and simplifying the resulting sum, I was able to show that
$$\int_1^{\sqrt{k+1}}\frac{1}{\lfloor x^2\rfloor}dx=\sum_{n=1}^{k} \frac{\sqrt{n+1}-\sqrt n}{n}$$
is true for every $k\geq 0$, which yields
$$\int_1^\infty \frac{1}{\lfloor x^2\rfloor}dx=\sum_{n=1}^\infty\frac{\sqrt{n+1}-\sqrt n}{n}$$
after letting $k\to\infty$. This equality is the main reason why I'm interested in the sum of the aforementioned series.
After some (unsurprisingly) futile attempts to evaluate the integral, I expect there to be no closed-form expression for the sum, which is why I'm open to an analytic expression (gamma function, Bessel functions, Riemann zeta function, etc.). Any help is appreciated.
Edit: after seeing the bounds provided by Markus Scheuer and Jorge, I thought I'd share some of my own.
From the fact that $x-1<\lfloor x\rfloor<x$ is true for every non-integer $x\geq 1$, we can infer that for every integer $k\geq 1$,
$$\int_\sqrt{k+1}^\infty \frac{1}{x^2}dx<\int_\sqrt{k+1}^\infty \frac{1}{\lfloor x^2\rfloor}dx<\int_\sqrt{k+1}^\infty \frac{1}{x^2-1}dx$$
Using
$$\int_{x}^{\infty}\frac{1}{t^2-1}dt=\coth^{-1}(x)$$
and
$$\int_\sqrt{k+1}^\infty\frac{1}{\lfloor x^2\rfloor}dx=\int_1^\infty\frac{1}{\lfloor x^2\rfloor}dx-\int_1^\sqrt{k+1}\frac{1}{\lfloor x^2\rfloor}dx=\sum_{n=1}^\infty \frac{\sqrt{n+1}-\sqrt n}{n}-\sum_{n=1}^{k}\frac{\sqrt{n+1}-\sqrt n}{n}$$
we deduce that
$$\frac{1}{\sqrt{k+1}}+\sum_{n=1}^{k}\frac{\sqrt{n+1}-\sqrt n}{n}<\sum_{n=1}^\infty\frac{\sqrt{n+1}-\sqrt n}{n}<\coth^{-1}\left(\sqrt{k+1}\right)+\sum_{n=1}^{k}\frac{\sqrt{n+1}-\sqrt n}{n}$$
 A: Complementing Markus Scheuer answer, an improvement of the lower bound with a nicer closed expresion can be achieved
\begin{align*}
\color{blue}{\sum_{n=1}^{\infty} \frac{\sqrt{n+1}-\sqrt{n}}{n}} > \sum_{n=1}^{\infty} \frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}\sqrt{n+1}}
= \sum_{n=1}^{\infty} \left( \frac{1}{\sqrt{n}} - \frac{1}{\sqrt{n+1}} \right)= \color{blue} 1
\end{align*}
The last equality follows because the defined series is a telescoping one with a simple limit
A: Hint: At least we have nice upper and lower bounds for the series.

We obtain by expanding with $\sqrt{n+1}+\sqrt{n}$:
\begin{align*}
\color{blue}{\sum_{n=1}^{\infty} \frac{\sqrt{n+1}-\sqrt{n}}{n}}
&=\sum_{n=1}^{\infty }\frac{1}{n\left(\sqrt{n+1}+\sqrt{n}\right)}\\
&>\sum_{n=1}^{\infty }\frac{1}{(n+1)\left(\sqrt{n+1}+\sqrt{n+1}\right)}\\
&= \frac{1}{2}\sum_{n=1}^{\infty }\frac{1}{(n+1)^{\frac{3}{2}}}\\
&=\frac{1}{2}\sum_{n=2}^\infty \frac{1}{n^{\frac{3}{2}}}\\
&\,\,\color{blue}{=\frac{1}{2}\zeta\left(\frac{3}{2}\right)-\frac{1}{2}}\\
\end{align*}
and on the other hand we have
\begin{align*}
\color{blue}{\sum_{n=1}^{\infty} \frac{\sqrt{n+1}-\sqrt{n}}{n}}
&=\sum_{n=1}^{\infty }\frac{1}{n\left(\sqrt{n+1}+\sqrt{n}\right)}\\
&<\sum_{n=1}^{\infty }\frac{1}{n\left(\sqrt{n}+\sqrt{n}\right)}\\
&= \frac{1}{2}\sum_{n=1}^{\infty }\frac{1}{n^{\frac{3}{2}}}\\
&\,\,\color{blue}{=\frac{1}{2}\zeta\left(\frac{3}{2}\right)}\\
\end{align*}

and since the sequence $\left(\sum_{n=1}^{N} \frac{\sqrt{n+1}-\sqrt{n}}{n}\right)_{1\leq N<\infty}$ of partial sums is strictly increasing and bounded above by $\frac{1}{2}\zeta\left(\frac{3}{2}\right)$ we know the series converges.
A: I do not know but to me it is so nice a candidate for Euler-Maclaurin. I promise no closed form just kind of close evaluation. When I write $\infty$ I mean $\lim\limits_{n \to \infty}$
$$\sum_{n=1}^{\infty}\frac{\sqrt{n+1}-\sqrt{n}}{n}=\int\limits_{1}^{\infty}\frac{\sqrt{x+1}-\sqrt{x}}{x}\, dx + \sqrt{2}-1 + D = 1 - \sqrt{2} + 2 \log(1+\sqrt{2}) + D$$
$$f(x)=\frac{\sqrt{x+1}-\sqrt{x}}{x}$$
$$D=-\sum_{k=1}^{\infty}\frac{B_{2k}}{(2k)!}f^{(2k-1)}(1)$$
Now it is the matter of finding the odd derivatives of $f(x)$ at $1$.
Take $u(x)=\sqrt{x+1}-\sqrt{x}$ and $v(x)=\frac{1}{x}$ and apply
$$(uv)^{(n)}=\sum_{k=0}^n {n \choose k} u^{(n-k)} v^{(k)}$$
since derivatives of $u$ and $v$ are straightforward and you have it
$$u^{(n)}(x)=(-1)^{n+1}\frac{(2n-3)!!}{2^n}(\frac1{(x+1)^{\frac{2n-1}{2}}}-\frac1{x^{\frac{2n-1}{2}}})$$
$$v^{(n)}(x)=(-1)^{n}n!\frac{1}{x^{n+1}}$$
Or
$$u^{(n)}(1)=(-1)^{n+1}\frac{(2n-3)!!}{2^n}(\frac1{2^{\frac{2n-1}{2}}}-1)$$
$$v^{(n)}(1)=(-1)^{n}n!$$
Finally
$$(uv)^{(n)}(1)=(-1)^{n+1}\sum_{k=0}^n {(n)_k}(2n-2k-3)!!(\frac{\sqrt{2}}{2^{2n-2k}}-\frac1{2^{n-k}})$$
$$\sum_{n=1}^{\infty}\frac{\sqrt{n+1}-\sqrt{n}}{n}=1 - \sqrt{2} + 2 \log(1+\sqrt{2})+$$ $$\sum_{p=1}^{\infty}(-1)^{2p+1}\frac{B_{2p}}{(2p)!}\sum_{k=0}^{2p-1} {(2p-1)_k}(4p-2k-5)!!(\frac{\sqrt{2}}{2^{4p-2k-2}}-\frac1{2^{2p-k-1}})$$
Not closed, but still calculable.
A: I'm also doubtful that a closed form exists, but you can get really good approximations using power series and the zeta function.
Note:
$$
\begin{align}
\sum \frac{\sqrt{n+1} - \sqrt{n}}{n} 
&= \sum \frac{\sqrt{n} \left ( \sqrt{1 + \frac{1}{n}} - 1 \right )}{n} \\\\
&= \sum \frac{1}{\sqrt{n}} \left ( \frac{1}{2n} - \frac{1}{8n^2} + \ldots \right ) \\\\
&= \frac{1}{2} \sum \frac{1}{n^{3/2}} - \frac{1}{8} \sum \frac{1}{n^{5/2}} + \ldots \\\\
&= \frac{1}{2} \zeta \left ( \frac{3}{2} \right ) - \frac{1}{8} \zeta \left ( \frac{5}{2} \right ) + \ldots
\end{align}
$$
Using the full series for $\sqrt{1+x}$, we find
$$ \sum_{n \geq 1} \frac{\sqrt{n+1} - \sqrt{n}}{n} = \sum_{k \geq 1} \binom{1/2}{k} \zeta \left ( \frac{2k+1}{2} \right ) .$$
Using series expansions for $\zeta \left ( \frac{2k+1}{2} \right )$ (which tends to $1$ as $k \to \infty$) and $\binom{1/2}{k}$ (which decays like $O \left ( k^{-3/2} \right )$) we can find a $k$ which gets us any desired precision.

I hope this helps ^_^
