Does the sum of the series $\sum_{n=1}^{\infty}\frac{\sqrt{n+1}-\sqrt n}{n}$ have an analytic expression?

Question

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

$$\frac{1}{\sqrt{k+1}}+\sum_{n=1}^{k}\frac{\sqrt{n+1}-\sqrt n}{n}<\sum_{n=1}^\infty\frac{\sqrt{n+1}-\sqrt n}{n}<\coth^{-1}\left(\sqrt{k+1}\right)+\sum_{n=1}^{k}\frac{\sqrt{n+1}-\sqrt n}{n}$$

Answer 1

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.

Answer 2

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

Answer 3

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.

Answer 4

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.

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I hope this helps ^_^