Given the sequence of real numbers $$a_n = \left(\sum_{k=1}^n \frac{1}{\sqrt{1+k^2}}\right)- \ln(n+\sqrt{1+n^2}) , n\in \Bbb{N}$$ Test its convergence.
What I did was to separate the sequence from the series and testing the convergence for each of them. Checking the $\lim_{n\to\infty}\ln(n+\sqrt{1+n^2})=\infty$ so the sequence diverges and then by the integral test I found out that $\sum_{k=1}^n \frac{1}{\sqrt{1+k^2}}$ diverges too. Now I don’t know how to go on and states if the sequence diverges or not. Have I used the wrong method?
I will prove that
By the monotone convergence theorem, 1 and 2 imply that $(a_n)_{n\in\mathbb{N}}$ is convergent.
Before proceeding, I note that $\ln(x+\sqrt{x^2+1})=\sinh^{-1}(x)$, so $a_n$ can be rewritten as$^{(*)}$ $$ a_n = \left(\sum_{k=1}^n \frac{1}{\sqrt{1+k^2}}\right)- \sinh^{-1}(n). \tag{1} $$
Proof of 1: Since $f(x):=\frac{1}{\sqrt{1+x^2}}$ is a decreasing function of $x$, one has $$ \frac{1}{\sqrt{1+(n+1)^2}}<\int_n^{n+1}\frac{dx}{\sqrt{1+x^2}}=\sinh^{-1}(n+1)-\sinh^{-1}(n), \tag{2} $$ hence $$ a_{n+1}-a_n=\frac{1}{\sqrt{1+(n+1)^2}}-\sinh^{-1}(n+1)+\sinh^{-1}(n)<0.\quad{\square} \tag{3} $$ Proof of 2: In a similar vein, one has $$ \sum_{k=1}^{n}\frac{1}{\sqrt{1+k^2}}>\int_1^{n+1}\frac{dx}{\sqrt{1+x^2}} =\sinh^{-1}(n+1)-\sinh^{-1}(1), \tag{4} $$ hence $$ a_n > \sinh^{-1}(n+1)-\sinh^{-1}(1)-\sinh^{-1}(n)>-\sinh^{-1}(1). \quad{\square} \tag{5} $$
$^{(*)}$ See, for instance, https://en.wikipedia.org/wiki/Hyperbolic_functions#Inverse_functions_as_logarithms.