5

How to compute $\sum_n (2n - \sqrt{n^2+1}-\sqrt{n^2-1})$? I tried two ways:

1. \begin{align*} (2n - \sqrt{n^2+1}-\sqrt{n^2-1}) &= n - \sqrt{n^2+1} + n -\sqrt{n^2-1} \\ &= \frac{1}{n+\sqrt{n^2-1}}-\frac{1}{n-\sqrt{n^2+1}}, \end{align*} but I don't know how to do later.

2. \begin{align*} (2n - \sqrt{n^2+1}-\sqrt{n^2-1}) &= 2n - \frac{(\sqrt{n^2+1} + \sqrt{n^2-1})}{1} \\ &= 2n - \frac{2}{\sqrt{n^2+1} - \sqrt{n^2-1}}, \end{align*} but I don't know how to do later too.

Sangchul Lee
  • 167,468
xxyshz
  • 795
  • 2
    The sum can be represented as $$\int_{0}^{\infty} \frac{I_1(x) - J_1(x)}{x(e^x-1)} , dx$$ where $J_1$ (resp. $I_n$) is the Bessel function (resp. modified Bessel function) of the first kind. I am not certain to whether this has a closed form. Also I edited your answer so that it reads better, hope this is to your liking. – Sangchul Lee Jun 16 '18 at 03:16
  • @SangchulLee thank you, but I think it may have simple ways to do since it's just a homework from calculus... – xxyshz Jun 16 '18 at 07:51
  • @SangchulLee. I don't know how you arrived to this integral. I am just unable to evaluate it numerically. May I ask what you get for it ? Thanks. – Claude Leibovici Jun 16 '18 at 08:22
  • 2
    I can tell you from my experience that finding its exact value is likely beyond our current knowledge, not just beyond the calculus level. Perhaps the correct problem (in calculus level) should only ask about its convergence. @ClaudeLeibovici, I obtained this from Laplace transform. But this representation seems inferior when it comes to numerical computation, as I also experienced Mathematica 11 fails to perform numerical integration on it. – Sangchul Lee Jun 16 '18 at 09:43
  • @SangchulLee: you might be interested in your approach producing a manageable integral representation, please see below. – Jack D'Aurizio Jun 16 '18 at 18:47

2 Answers2

2

Starting with Sangchul Lee's integral representation (which is a consequence of the Laplace transform) $$ S = \int_{0}^{+\infty}\frac{I_1(x)-J_1(x)}{x(e^x-1)}\,dx = \frac{1}{\pi}\int_{0}^{+\infty}\int_{0}^{\pi}\frac{e^{-x\cos\theta}-e^{ix\cos\theta}}{e^x-1}\sin^2\theta\,d\theta\,dx \tag{1}$$ and applying Fubini's theorem we get $$ S = \frac{1}{\pi}\int_{0}^{\pi}\left[\psi(1-i\cos\theta)-\psi(1+\cos\theta)\right]\sin^2\theta\,d\theta \tag{2}$$ where $$\begin{eqnarray*} S &=& \frac{1}{2\pi}\int_{0}^{\pi}\left[\psi(1-i\cos\theta)+\psi(1+i\cos\theta)-2\psi(1-\cos\theta)\right]\sin^2\theta\,d\theta\\&=&\frac{1}{\pi}\int_{0}^{\pi}\left[\log\Gamma(1-\cos\theta)-\text{Im}\,\log\Gamma(1-i\cos\theta)\right]\cos\theta\,d\theta \tag{3}\end{eqnarray*}$$ allows an efficient numerical evaluation of $S$ through standard integration techniques (composite Simpson's rule or Gaussian quadrature): $$ S \approx 0.6369740582412\tag{4} $$ but I do not believe that $S$ has a simple closed form in terms of standard mathematical constants.
We have a similar situation here.

Jack D'Aurizio
  • 353,855
0

Let us assume that you need to compute the infinite summation with $$a_n=2n - \sqrt{n^2+1}-\sqrt{n^2-1}$$ For large values of $n$, rewrite $$a_n=n\left(2- \sqrt{1+\frac 1{n^2}}- \sqrt{1-\frac 1{n^2}}\right)$$ and use the binomial expansion or Taylor series to get $$a_n=\frac{1}{4 n^3}+\frac{5}{64 n^7}+O\left(\frac{1}{n^{11}}\right)$$ So, $$\sum_{n=1}^\infty a_n\approx\sum_{n=1}^p a_n+\frac 14 \sum_{n=p+1}^\infty \frac{1}{ n^3}=\sum_{n=1}^p a_n-\frac{1}{8}\psi ^{(2)}(p+1)\approx \sum_{n=1}^p a_n +\frac 1 {8 p^2}$$

If we use $p=10$, the first summation is $\approx 0.635843$, the correction term is $0.00125$ making a total of $0.637093$ while the infinite summation looks to be $\approx 0.636974$. For sure, if we increase $p$, we shall get closer and closer.

Now, the question is : what is this number ?

Claude Leibovici
  • 260,315