Integral Representation of $F(n)$

Question

Is there a nice integral representation of

$$F(n) = \sum_{k = 0}^{\infty}\left(\sqrt{k + 1} - \sqrt{k}\right)^n$$

where $n \in \mathbb{N}$ ? I'm working on this problem for a long time. I was able to find a weird integral representation for $n = 3$ and $n = 4$, however I'm wondering if there's a nice approach to find the same (integral representation or closed form) for any $n \in \mathbb{N}$. I'm interested in odd values of $n$.

It would be nice if the representation is in single integral but there's nothing wrong with complicated integrals or closed forms. Thanks.

EDIT $\textbf{1}$: For $n = 3$, we have

$$\sum_{k = 0}^{\infty}\left(\sqrt{k + 1} - \sqrt{k}\right)^3 = \frac{3}{\sqrt{\pi}}\int_{0}^{+\infty}\left[\frac{1}{2x^{3/2}e^x}-\frac{1}{x^{5/2}e^x}+\frac{1}{x^{5/2}e^x(e^x-1)}\right]\,dx$$

For $n = 4$, we have

$$\sum_{k = 0}^{\infty}\left(\sqrt{k + 1} - \sqrt{k}\right)^4 =\frac{4\pi}{3}\int_{0}^{1}\frac{x^{3/2}(1-x)^{3/2}}{\sin^2(\pi x)}\,dx$$

Answer 1

If $n=2m+1\ge 3$ is odd then it has a "closed form" in term of $\zeta$ values: $$((k+1)^{1/2}-k^{1/2})^n=\sum_{j=0}^m a_j(n) k^{1/2+j}+b_j(n) (k+1)^{1/2+j}$$ so that continuing analytically to $\Re(s) \ge 0$ $$\sum_{k\ge 1}\left(\sum_{j=0}^m a_j(n) k^{1/2+j-s}+b_j(n) (k+1)^{1/2+j} (k+1)^{-s}\sum_{l=0}^m {-s\choose l}(-1)^l (k+1)^{-l}\right) $$ we obtain $$\sum_{k\ge 1} ((k+1)^{1/2}-k^{1/2})^n=\sum_{j=0}^m (a_j(n)+b_j(n)) \zeta(-1/2-j)-\sum_{j=0}^m b_j(n) $$ Then pick your favorite integral representation of $\zeta(s)$.

For $n$ even the same idea gives that your series is a sum of $\sum_{k\ge 1} k^{1-s}$ and $\sum_{k\ge 1} (k(k+1))^{1/2} k^{-s}$ at negative integers and you already know how to obtain an integral representation for the latter.

Answer 2

If $\;z\in(1,\infty),\;$ then the function $$f_n(z)=\dfrac\pi4\,(z^2-1)\,z^{\large-\frac{n+4}2}\cot{\pi\dfrac{(z-1)^2}{4z}}\tag1$$ has the poles in the points $$z_k = 2k+1 +2\sqrt{k^2+k} =\left(\sqrt{k+1}+\sqrt k\right)^2,\quad k\in\mathbb N,\tag2$$ wherein $$\underset{\large {z_k}\small^\mathstrut}{\operatorname{Res}}f_n(z) =\dfrac\pi4\,(z^2_k-1)z_k^{\large-\frac{n+4}2}\,{\cos\pi\dfrac{(z-1)^2}{4z}} \lim\limits_{\large z\to {z_k}\Large\mathstrut}\,\dfrac{z-z_k}{\sin\pi\dfrac{(z-1)^2}{4z}} $$$$ =\dfrac\pi4\,(z^2_k-1)z_k^{\large-\frac{n+4}2}\,{\cos\pi\dfrac{(z-1)^2}{4z}} \lim\limits_{\large z\to {z_k}\Large\mathstrut}\, \dfrac1{\dfrac\pi4\left(1-\dfrac1{z^2}\right)\cos\pi\dfrac{(z-1)^2}{4z}} = z_k^{\large-\frac n2},$$ $$\underset{\large {z_k}\small^\mathstrut}{\operatorname{Res}}f_n(z) = \left(\sqrt{k+1} - \sqrt k\right)^n.\tag3$$

Choosen the arbitrary infinity contour $\;C\;$ with the points $\;z_1, z_2,\dots,\;$ one can get $$S_n=\sum\limits_{k=0}^\infty\left(\sqrt{k+1} - \sqrt k\right)^n = 1+\dfrac1{2\pi i}\oint\limits_C f_n(z)\text dz\tag4$$

In particular, for the contours $$C =\left\{\dfrac{1}{1-t^2}+p+32it,\;t\in(-1,1),\;p\in[1,4]\right\}$$

easily to get the integral representation in the form of $$S_n= 1+\dfrac1{2\pi i}\int\limits_{-1}^1 f_n\left(\dfrac{1}{1-t^2}+p +32it\right)\left(\dfrac{2t}{(1-t^2)^2}+32i\right)\text dt.\tag5$$

On the other hand, $$\left(\sqrt{k+1}-\sqrt k\right)^n = \left(2k+1+\sqrt{(2k+1)^2-1}\right)^{\large-\frac n2} $$ $$= (2k+1)^{\large-\frac n2}\left(1+\sqrt{1-\dfrac1{(2k+1)^2}}\right)^{\large-\frac n2}$$ $$= (4k+2)^{\large-\frac n2}\left(1-\dfrac1{(4k+2)^2}-\dfrac1{(4k+2)^4}-\dots\right)^{\large-\frac n2}$$ $$= (4k+2)^{\large-\frac n2}\left(1+\dfrac n{2(4k+2)^2}+\dfrac n{2(4k+2)^4} +\dfrac{n(n+2)}{8(4k+2)^4}+\dots\right),$$ $$\begin{align} &S_n\approx \sum\limits_{k=0}^{s-1}\left(\sqrt{k+1} + \sqrt k\right)^{-n} + 2^{-n}\\[4pt] &\times\left(\zeta\left(\dfrac n2, s+\dfrac12\right) +\dfrac n{32}\zeta\left(\dfrac n2+2, s+\dfrac12\right) +\dfrac{n(n+6)}{2048}\zeta\left(\dfrac n2+4, s+\dfrac12\right) \right), \end{align}\tag6$$ where $\;\zeta(z,a)\;$ is the generalized Riemann zeta function.

Numeric calculations of the given sum via $(5)$ (with the same results for all values of $\;p\;$ from domain) and via $(6)$ for $\;s=35\;$ by Wolfram Alpha are shown in the table $(7).$

\begin{vmatrix} n & (5) & (6)\\ 3 & 1.24731\,73498\,64127\,39610\,384 & 1.24731\,73498\,64124\,19\\ 4 & 1.06038\,21964\,75434\,10179\,32212 & 1.06038\,21964\,75433\,7130\\ 5 & 1.01940\,80755\,93321\,43798\,17195 & 1.01940\,80755\,93321\,3941\\ 6 & 1.00693\,53143\,70406\,29845\,53855\,17 & 1.00693\,53143\,70406\,29373\,3\\ 7 & 1.00261\,51034\,44489\,88275\,39751\,41 & 1.00261\,51034\,44489\,88226\,2\\ 8 & 1.00101\,71884\,92355\,62561\,09186\,051 & 1.00101\,71884\,92355\,62556\,09\\ 9 & 1.00040\,33950\,87623\,01165\,83908\,220 & 1.00040\,33950\,87623\,01165\,34\\ 10 & 1.00016\,20240\,77476\,98571\,45268\,798 & 1.00016\,20240\,77476\,98571\,404\\ 11 & 1.00006\,56411\,98162\,21005\,94527\,8947 & 1.00006\,56411\,98162\,21005\,9406\\ 12 & 1.00002\,67536\,86676\,80665\,73794\,1985 & 1.00002\,67536\,86676\,80665\,73732\tag7 \end{vmatrix}

Therefore, the calculations confirm obtained formulas $(4)-(5).$

Answer 3

To start with, this will use an unconventional strategy of the definition of the definite integral as a Riemann sum with the number of subdivisions being the upper integration bound: $$\int_a^b f(x)dx=\lim_{N\to ∞}\sum_{k=0}^N f\biggl(a+k\frac{b-a}{N}\biggl )\frac{b-a}{N}$$

Then, using the above definition, let there be a function f such that: $$\int_a^b f(x)dx= \int_0^N f(x)dx= \int_0^N (\sqrt{x+1}-\sqrt x)^n dx$$

The integral and summation switch should be able to be done. As Combining the two steps together yields with n as a constant to x: $$\int_0^N (\sqrt{x+1}-\sqrt x)^n dx= \lim_{N\to ∞}\sum_{k=0}^N f\biggl(a+k\frac{b-a}{N}\biggl )\frac{b-a}{N}= \lim_{N\to ∞}\sum_{k=0}^N f\biggl(0+k\frac{N-0}{N}\biggl )\frac{N-0}{N}= \lim_{N\to ∞}\sum_{k=0}^N f(k)*1= \lim_{N\to ∞}\sum_{k=0}^N (\sqrt{k+1}-\sqrt k)^n= \sum_{k=0}^ ∞(\sqrt{k+1}-\sqrt k)^n$$

Which is your exact summation. Adding on, this integral can be expanded using the binomial theorem because n$\in \Bbb N$: $$F(n)=\sum_{k=0}^ ∞(\sqrt{k+1}-\sqrt k)^n= \int_0^N (\sqrt{x+1}-\sqrt x)^ndx= \int_0^N\sum_{m=0}^n\binom{n}{m}(k+1)^{\frac{n-m}{2}}(-1)^n(k)^{\frac{n}{2}} dx=\sum_{m=0}^n (-1)^n \binom{n}{m}\int_0^N(k+1)^{\frac{n}{2}}k^{\frac{n-m}{2}}dx$$

If n was negative, then this works perfectly:demo

Also, take a look at what happens for slightly larger two digit n such that the upper integral bound is equal to the summation for a particular n. They converge to almost identical values Demo 2

This means that a better definition is that:

$$F(n)=\sum_{k=0}^ ∞(\sqrt{k+1}-\sqrt k)^n= \int_0^{F(n)} (\sqrt{x+1}-\sqrt x)^ndx= \int_0^{\sum_{k=0}^ ∞(\sqrt{k+1}-\sqrt k)^n} (\sqrt{x+1}-\sqrt x)^ndx= \sum_{m=0}^n(-1)^n \binom{n}{m}\int_0^{\sum_{k=0}^ ∞(\sqrt{k+1}-\sqrt k)^n}(k+1)^{\frac{n}{2}}k^{\frac{n-m}{2}}dx$$

This inner integral can be expressed by the incomplete beta function:

$$B_z(a,b)=\int_0^zt^{a-1}(1-t)^{b-1} dt⇒(-1)^{\frac{n-m}{2}+1}\int_0^{-F(n)} x^{\frac {n-m}2}(x+1)^{\frac {n}{2}}(-dx)= (-1)^{\frac{n-m}{2}} B_{-F(n)}\bigl(\frac {n-m}2+1,\frac {n}2+1\bigl)=^{x=-t} \int_0^{F(n)}t^{(\frac {n-m}2+1)-1}(1-t)^{(\frac {n}{2} +1)-1}dt$$

The restricions on the beta function so that it converges are Re(a,b)>0, |z=-F(n)|<1. The real part restriction can be ignored because a and b are functions of $n,m\in \Bbb N$ as they are sum indices. Also, the sum may be the correct value, but negative, so absolute value bars are needed with the alternating part being able to be set to |-1|=1 given that $k,m,n\in \Bbb N$:

If the two previous steps are combined, then the final result is $$F(n)= \int_0^{F(n)} (\sqrt{x+1}-\sqrt x)^ndx= \int_0^{\sum_{k=0}^ ∞(\sqrt{k+1}-\sqrt k)^n} (\sqrt{x+1}-\sqrt x)^ndx=\biggl| \sum_{m=0}^n(-1)^{\frac{n-m}{2}+n}\binom{n}{m}\int_0^{F(n)}(k+1)^{\frac{n}{2}}k^{\frac{n-m}{2}}dx\biggl|= \biggl|\sum_{m=0}^n(-1)^{n+1} \binom{n}{m} B_{-F(n)}\bigl(\frac {n}2+1,\frac {n-m}2+1\bigl)\biggr|=$$ $$\biggr|\sum_{m=0}^n \binom{n}{m} B_{-\sum_{k=0}^ ∞(\sqrt{k+1}-\sqrt k)^n}\bigl(\frac n2+1,\frac {n-m}2+1\bigl)\biggr|, 0<-F(n)<1$$

As a last message, please click the demo 2 link if all of this seems nonsensical. This technique really works, and does as well for your wanted odd n. In particular, the values $n\ge 11$ function quite well. I decided it would be best to just include the incomplete beta function as this derivation is easiest to understand. Just pick any integral representation of the beta function and plug in the arguments making sure that the restrictions are met. The restriction at the end has the subscript function never reaching zero and having to be less than one to converge. Here is the demo for the expression with the binimial series corrected:Demo 3, you can see a similar decimal expansion. As always please give me feedback and correct me!