Does a closed form expression exist for
$$\sum_{k=1}^{n}\left(\sqrt{1+\frac{1}{k}}-\sqrt{1 -\frac{1}{k}}\right)$$
I obtained
$$0.97423066\ln(n)+1.2019463$$
using log regression and it works very well for my (physics) problems but I was wondering if a better solution exists. Is the Euler-Maclaurin method something to look at ?
Many thanks.
Since \begin{align*} \sqrt{1+\frac1k} - \sqrt{1-\frac1k} &= \frac{(\sqrt{1+\frac1k} - \sqrt{1-\frac1k})(\sqrt{1+\frac1k} + \sqrt{1-\frac1k})}{\sqrt{1+\frac1k} + \sqrt{1-\frac1k}} \\ &= \frac{\frac2k}{\sqrt{1+\frac1k} + \sqrt{1-\frac1k}} \\ &= \frac{\frac2k}{(1+O(\frac1k))+(1+O(\frac1k))} = \frac1k + O\biggl( \frac1{k^2} \biggr), \end{align*} the sum will be asymptotically $$ \sum_{k=1}^n \biggl( \sqrt{1+\frac1k} - \sqrt{1-\frac1k} \biggr) = \sum_{k=1}^n \biggl( \frac1k + O\biggl( \frac1{k^2} \biggr) \biggr) = \ln n + O(1) $$ with leading coefficient $1$. I doubt there's a particularly nice form for the constant other than $$ \gamma + \sum_{k=1}^\infty \biggl( \sqrt{1+\frac1k} - \sqrt{1-\frac1k} - \frac1k \biggr) $$ (where $\gamma$ is Euler's constant); but it turns out this sum converges as fast as $\sum \frac1{k^3}$, so even just the first $10^5$ terms already gives $1.01902854$ accurately rounded to that many decimal places.
I shall derive a complete asymptotic expansion for the sum as $n\to+\infty$. Note that for $|x|<1$, $$ \sqrt {1 + x} - \sqrt {1 - x} - x = \sum\limits_{m = 1}^\infty \binom{4m}{2m} \frac{{x^{2m + 1} }}{{16^m (2m + 1)}}. $$ Employing this expansion together with $$ \sum\limits_{k = 1}^n {\frac{1}{k}} = H_n ,\qquad \sum\limits_{k = 1}^n {\frac{1}{{k^{2m + 1} }}} = \zeta (2m + 1) - \sum\limits_{k = n + 1}^\infty {\frac{1}{{k^{2m + 1} }}} , $$ where $H_n$ is the $n$th harmonic number and $\zeta$ is the Riemann zeta function, we find $$ \sum\limits_{k = 1}^n {\left( {\sqrt {1 + \frac{1}{k}} - \sqrt {1 - \frac{1}{k}} } \right)} = H_n - \gamma + C - \sum\limits_{m = 1}^\infty \binom{4m}{2m} \frac{1}{{16^m (2m + 1)}}\sum\limits_{k = n + 1}^\infty {\frac{1}{{k^{2m + 1} }}} . $$ Here $$ C = \gamma + \sum\limits_{m = 1}^\infty \binom{4m}{2m} \frac{{\zeta (2m + 1)}}{{16^m (2m + 1)}}=1.019028540417608960\ldots, $$ and $\gamma$ is the Euler–Mascheroni constant. It is shown in this paper that $$ - \sum\limits_{k = n + 1}^\infty {\frac{1}{{k^{2m + 1} }}} \sim \frac{1}{{\left( {n + \frac{1}{2}} \right)^{2m} }}\sum\limits_{k = 0}^\infty {\frac{{(1 - 2^{1 - 2k} )B_{2k} }}{{(2k)!}}\frac{{ (2m + 2k-1)!}}{{(2m)!}}\frac{1}{{\left( {n + \frac{1}{2}} \right)^{2k} }}} $$ for any $m\ge 1$ as $n\to+\infty$. Here the $B_k$ denote the Bernoulli numbers. Substitution and re-arrangement yield \begin{multline*} - \sum\limits_{m = 1}^\infty \binom{4m}{2m} \frac{1}{{16^m (2m + 1)}}\sum\limits_{k = n + 1}^\infty {\frac{1}{{k^{2m + 1} }}} \\ \sim \sum\limits_{k = 1}^\infty {\left( {\frac{1}{{2k(2k + 1)}}\sum\limits_{m = 1}^k {\binom{4m}{2m}\binom{2k+1}{2m+1}\frac{{(1 - 2^{1 - 2k + 2m} )B_{2k - 2m} }}{{16^m }}} } \right)\frac{1}{{\left( {n + \frac{1}{2}} \right)^{2k} }}} , \end{multline*} as $n\to+\infty$. It is shown in this paper that $$ H_n - \gamma \sim \log \left( {n + \tfrac{1}{2}} \right) + \sum\limits_{k = 1}^\infty {\frac{{(1 - 2^{1 - 2k} )B_{2k} }}{{2k\left( {n + \frac{1}{2}} \right)^{2k} }}} , $$ as $n\to+\infty$. Hence, in summary, \begin{align*} \sum\limits_{k = 1}^n {\left( {\sqrt {1 + \frac{1}{k}} - \sqrt {1 - \frac{1}{k}} } \right)} & \sim \log \left( {n + \tfrac{1}{2}} \right) + C + \sum\limits_{k = 1}^\infty {\frac{{A_k }}{{\left( {n + \frac{1}{2}} \right)^{2k} }}} \\ & = \log \left( {n + \tfrac{1}{2}} \right) + C - \frac{1}{{48\left( {n + \frac{1}{2}} \right)^2 }} - \frac{{41}}{{7680\left( {n + \frac{1}{2}} \right)^4 }} + \ldots , \end{align*} as $n\to+\infty$, with $$ A_k = \frac{1}{{2k(2k + 1)}}\sum\limits_{m = 0}^k {\binom{4m}{2m}\binom{2k+1}{2m+1}\frac{{(1 - 2^{1 - 2k + 2m} )B_{2k - 2m} }}{{16^m }}} . $$