Is $S_n=\frac{1}{\sqrt{n}}\sum_{j=1}^{n-1}\frac{1}{\sqrt{j+1}}$ convergent? Suppose $$S_n=\frac{1}{\sqrt{n}}\sum_{j=1}^{n-1}\frac{1}{\sqrt{j+1}},{\space}n\geq2$$ 
Is this series convergent as $n\to\infty$?
$\mathbf{My{\space}Tries}$
The first idea comes to my mind is to prove that this sequence is actually a Cauchy  Sequence, however it failed because I found that (just skip the calculation)$$|S_m-S_n|<\sqrt{\frac{m}{n-1}}$$
And then I'm stuck here, since I can not find a upper bound for $\sqrt{\frac{m}{n-1}}$.
Then I resorted to some other convergence tests, like the Ratio test. However, I found these tests are not necessarily available in this situation because as $n\to\infty$, the term $\frac{1}{\sqrt{n}}$ also affects the behavior of $S_n$ and $\frac{\infty}{\infty}$ is possible.
Finally, I just simulated this series on R and it seems to be convergent. Is there any rigorous idea that can be applied to prove this?
Thank you!
 A: Basically the integral test for convergence. 
$$\frac{2(\sqrt{n+1}-\sqrt{2})}{\sqrt{n}}=\frac{1}{\sqrt{n}}\int_{2}^{n+1}\frac{1}{\sqrt{x}}dx\leq S_n\leq \frac{1}{\sqrt{n}}\int_{1}^{n}\frac{1}{\sqrt{x}}dx=\frac{2(\sqrt{n}-\sqrt{1})}{\sqrt{n}}$$
Taking limits we get $\lim_{n\to\infty}S_n=2$. 
A: By Stolz theorem $$\lim\limits_{n\to\infty}S_n=\lim\limits_{n\to\infty}\frac{\sum\limits_{j=1}^n\frac{1}{\sqrt{j+1}}-\sum\limits_{j=1}^{n-1}\frac{1}{\sqrt{j+1}}}{\sqrt{n+1}-\sqrt{n}}=\frac{\frac{1}{\sqrt{n+1}}}{\sqrt{n+1}-\sqrt{n}}=\frac{\sqrt{n+1}+\sqrt{n}}{\sqrt{n+1}}=2.$$
A: Since you already received good answers about the limit, let me show that we can have quite good approximations of the partial terms.
If you are familiar with generalized harmonic numbers, we have
$$\sqrt{n}\,S_n=\sum_{j=1}^{n-1}\frac{1}{\sqrt{j+1}}=H_n^{\left(\frac{1}{2}\right)}-1$$ Using the asymptotics of the rhs would then give
$$S_n=2+ \left(\zeta \left(\frac{1}{2}\right)-1\right)\frac{1}{\sqrt{n}}+\frac{1}{2
   n}-\frac{1}{24 n^2}+O\left(\frac{1}{n^4}\right)$$ which shows the limit and how it is approached. 
Using for $n=10$, the exact value would be $1.2715512$ while the above truncated expansion would give $1.2715509$.
