How to calculate $\lim_{n\to\infty}\left(n-\sum_{k=2}^n\frac{k}{\sqrt{k^2-1}} \right)$? This is an admission question of Tokyo University：
For all natural number $n\geqslant2$, we always have
$$n-\sum_{k=2}^n\frac{k}{\sqrt{k^2-1}}\geqslant\cfrac{i}{10}$$
in which $i$ is integer, please calculate the maximum 0f $i$.
I have worked out this problem, furthermore, how to calculate
$$\lim_{n\to\infty}\left(n-\sum_{k=2}^n\frac{k}{\sqrt{k^2-1}}\right)$$
is the value a transcendental number？
 A: First note that
$$
\frac{k}{{\sqrt {k^2  - 1} }} = 1 + \frac{1}{{(k + \sqrt {k^2  - 1} )\sqrt {k^2  - 1} }}.
$$
Thus,
$$
\mathop {\lim }\limits_{n \to  + \infty } \left( {n - \sum\limits_{k = 2}^n {\frac{k}{{\sqrt {k^2  - 1} }}} } \right) = \mathop {\lim }\limits_{n \to  + \infty } \left( {1 - \sum\limits_{k = 2}^n {\frac{1}{{(k + \sqrt {k^2  - 1} )\sqrt {k^2  - 1} }}} } \right).
$$
Observe that our sequence is decreasing, so we indeed need to estimate this limit efficiently. Now
$$
\sum\limits_{k = 2}^n {\frac{1}{{(k + \sqrt {k^2  - 1} )\sqrt {k^2  - 1} }}}  \le \frac{1}{2}\sum\limits_{k = 2}^n {\frac{1}{{k^2  - 1}}}  \to \frac{1}{2}\sum\limits_{k = 2}^\infty  {\frac{1}{{k^2  - 1}}}  = \frac{3}{8}
$$
and
$$
\sum\limits_{k = 2}^n {\frac{1}{{(k + \sqrt {k^2  - 1} )\sqrt {k^2  - 1} }}}  \ge \frac{1}{2}\sum\limits_{k = 2}^n {\frac{1}{{k^2 }}}  \to \frac{1}{2}\sum\limits_{k = 2}^\infty  {\frac{1}{{k^2 }}}  = \frac{1}{2}\left( {\frac{{\pi ^2 }}{6} - 1} \right) \ge 0.322.
$$
Consequently,
$$
0.625 \le \mathop {\lim }\limits_{n \to  + \infty } \left( {n - \sum\limits_{k = 2}^n {\frac{k}{{\sqrt {k^2  - 1} }}} } \right) \le 0.678,
$$
and the answer is $i=6$.
