Prove convergence of two series: I would like to prove if the next two series are convergent. 
First:
$$
\sum_{n=1}^{\infty}\log\left(\frac{n+1}{n}\right)\arcsin \left(\frac{1}{\sqrt{n}}\right)
$$
I think that this series is convergent, so $$\arcsin\left(\frac{1}{\sqrt{n}}\right)$$ is similar to $$\frac{1}{\sqrt {n}}$$. And $$\log\left(\frac{n+1}{n}\right)=\log\left(1+\frac{1}{n}\right)\sim \frac {1}{n} $$ if n goes to infinity. 
So I have the series $$\sum_{n=1}^{\infty}\frac{1}{n}\frac{1}{\sqrt {n}}$$, this series converge. Is this argument valid to prove the convergence?
Second: 
$$\sum_{n=1}^{\infty}1-\sec\left(\frac{1}{n}\right)$$.
Could you help me please? Give me some clue please!!!!
Thank you

Answer for the second series: this series converges absolutely  if $\sum \frac {1-\cos(\frac 1 n)} {|\cos(\frac 1 n)|}$ converges. Since the denominator tends to $1$ it is enough to prove convergence of $\sum  {(1-\cos(\frac 1 n))}$. This series converges because $1-\cos \theta \leq \frac {\theta^{2}} {2}$ and $\sum \frac 1 {n^{2}} <\infty$. [ Proof of $1-\cos \theta \leq \frac {\theta^{2}} {2}$: $1-\cos \theta =2\sin^{2} \frac {\theta} 2 \leq \frac {\theta^{2}} {2}$ since $|\sin t| \leq t$ for all $t \geq 0$]. 
