How to compute $\int\limits_{1}^{\infty}{(\sqrt[x]{x+1}-\sqrt[x]{x-1})}\,dx$?

Question

I was trying to prove that the series $\sum\limits_{n=1}^{\infty}{(\sqrt[n]{n+1}-\sqrt[n]{n-1})}$ converges, in a way that is different to proving convergence by limit-comparison test, that is, showing that $\lim\limits_{n\to\infty}{\left(\frac{\sqrt[n]{n+1}-\sqrt[n]{n-1}}{n^{-2}}\right)} = 2$, as in this post of mine. That is because in an examination setting, attempting to check for convergence with possible candidates by trial and error can take a lot of time. So, I wanted to take the seemingly straight-forward route - the integral test.
Needless to say, the definite integral $\int\limits_{1}^{\infty}{(\sqrt[x]{x+1}-\sqrt[x]{x-1})}\,dx$ converges. So, if I were to prove the convergence of the series using the integral test on paper, I could only think of first computing the indefinite integral $\int{(\sqrt[x]{x+1}-\sqrt[x]{x-1})}\,dx$ and then applying the limits. But despite investing substantial time into computing a simpler integral - $\int{(\sqrt[x]{x})}\,dx$, I couldn't do it, but neither could WolframAlpha. While I am aware of the fact that there are other tests that I could employ, I really want to know the solution to this integral. If it's the case that the indefinite integral does not exist, is it possible to compute the definite integral by pen-and-paper? If so, can anyone please show me how?

Answer 1

If you are optimistic, for large values of $x$, by Taylor expansion

$$\sqrt[x]{x+1}-\sqrt[x]{x-1}=\frac 2 {x^2}\sum_{n=0}^\infty \frac{P_n(L)} {n! \,x^n}\qquad \text{where}\qquad L=\log(x)$$ where the first polynomials are $$\left( \begin{array}{cc} n & P_n(L) \\ 0 & 1 \\ 1 & L \\ 2 & L^2+\frac{2}{3} \\ 3 & L^3+2 L-3 \\ 4 & L^4+4 L^2-12 L+\frac{44}{5} \\ 5 & L^5+\frac{20 }{3}L^3-30 L^2+44 L-50\\ 6 & L^6+10 L^4-60 L^3+132 L^2-300 L+\frac{2190}{7}\\ 7 & L^7+14 L^5-105 L^4+308 L^3-1050 L^2+2190 L-2184\\ \end{array} \right)$$

We shall use the fact that, for $m>0$, $n>1$ and $n>m$

$$I_{m,n}=\int_1^\infty \frac{\log^m(x)}{x^n}\,dx=\frac {m!}{(n-1)^{m+1}}$$

Computing the partial sums, this leads to the sequence

$$\left\{2,\frac{5}{2},\frac{151}{54},\frac{8971}{3456},\frac{29250887}{10800000}, \frac{754026449}{291600000},\frac{647126560028807}{240145138800000},\frac{4089558 9925373992427}{15738151816396800000},\frac{22474001837129448365077307}{83638991 39457731788800000}\right\}$$ which is oscillating but converges (rather slowly) to the value obtained by numerical integration $(2.64415)$.

What could be interesting to notice is the behaviour of the average of two consecutive terms $$\left( \begin{array}{cc} n & \frac 12 (R_n+R_{n-1}) \\ 3 & 2.25000 \\ 4 & 2.64815 \\ 5 & 2.69604 \\ 6 & 2.65210 \\ 7 & 2.64712 \\ 8 & 2.64028 \\ 9 & 2.64662 \\ 10 & 2.64276 \\ 11 & 2.64627 \\ 12 & 2.64312 \\ 13 & 2.64565 \\ 14 & 2.64331 \\ 15 & 2.64527 \\ 16 & 2.64347 \\ 17 & 2.64502 \\ 18 & 2.64358 \\ 19 & 2.64484 \\ 20 & 2.64367 \\ \end{array} \right)$$

Using the exact numbers and acceleration, the obtained extrapolated limit is $\color{red}{2.6441}752$ to be compared to $\color{blue}{2.6441595}$ (absolute error of $2.86\times 10^{-5}$).