How to show that $a_n=1+1/\sqrt{2}+\cdots+(1/\sqrt{n-1})-2\sqrt{n}$ has an upper bound. Let $a_n=1+1/\sqrt{2}+\cdots+(1/\sqrt{n-1})-2\sqrt{n}$ while   $a_1=-2,n\ge2 $ ,
I need to prove that $a_n$ converges.
I proved that it is monotonically increasing and tried to prove that it is upper-bounded by induction but failed to.
Also, it was told that $a_n$ converges to $-2<L<-1$ so I tried to show by induction that $a_n$ is bounded by $-1$, but I'm always stuck with $a_{n+1} \le -1 + 1/\sqrt{n}$ or something like that.
How can I prove that $a_n$ is bounded from above?

Note: To prove something converges you don't have to figure out what it converges to.  And to prove something is bounded above you don't have to find the least upper bound; it's enough just to find any upper bound (and prove it is an upper bound).
So
$a_{n+1} - a_n = \frac 1{\sqrt{n}} - 2\sqrt{n+1} + 2\sqrt{n}$.  
Claim: $2\sqrt n + \frac 1{\sqrt n} > 2\sqrt{n+1}$ for all natural $n$.
Pf: As all terms are positive and greater than zero...
$2\sqrt n + \frac 1{\sqrt n} > 2\sqrt{n+1} \iff$
$(2\sqrt n + \frac 1{\sqrt n})^2 > (2\sqrt{n+1})^2 \iff $
$4 n + 4 + \frac 1n > 4(n+1) \iff$
$\frac 1n > 0$.
So it it is true.
And so $a_{n+1} - a_n = 2\sqrt{n} + \frac 1{\sqrt n} - 2\sqrt{n+1} > 0$ so $a_{n+1} > a_n$.
So $\{a_i\}$ is monotonically increasing.
Claim:  $a_{n} \le -1 + \frac 1{\sqrt{n-1}} \le 0$ for all $n\ge 2$.
Base case: $a_2 = 1 - 2\sqrt{2} \le 1-2 = -1 < -1 + \frac 1{\sqrt{1}} \le 0$.
Inductive step:
If $a_n \le -1 +\frac 1{\sqrt{n-1}}$ then 
$a_{n+1} = a_n + \frac 1{\sqrt{n}} - 2\sqrt{n} + 2\sqrt{n-1}$
$\le -1 +\frac 1{\sqrt{n-1}}+ \frac 1{\sqrt{n}} - 2\sqrt{n} + 2\sqrt{n-1}$
We proved above that $\frac 1{\sqrt{n-1}}- 2\sqrt{n} + 2\sqrt{n-1} > 0$ so 
$a_{n+1} \le  -1 +\frac 1{\sqrt{n-1}}+ \frac 1{\sqrt{n}} - 2\sqrt{n} + 2\sqrt{n-1} < -1 +\frac 1{\sqrt{n}} < -1 + 1 = 0$.
So the claim is true.
$a_n \le -1 + \frac 1{\sqrt{n-1}} \le 0$ so $\{a_i\}$ is bounded above by $0$.
Claim 2:  $\{a_i\}$ is bounded above by $-1$.
Claim:  Let $\epsilon > 0$.  Then $\frac 1{\sqrt{n}} < \epsilon \iff n > \frac 1{\epsilon^2}$.
So for all  $n + 1 > \frac 1{\epsilon^2}$ then $a_{n+1} \le -1 + \frac 1{\sqrt{n}} < - 1 + \epsilon$.  But as $\{a_i\}$ is monotonically increasing.  If $m \le n \le \frac 1{\epsilon^2} -1$ then $a_m < a_{n+1} < -1 + \epsilon$ and if $n \ge n+1 >  \frac 1{\epsilon^2}$ then $a_m \le -1 + \frac 1{\sqrt{m-1}} < -1 + \epsilon$.
So all $a_i < 1 + \epsilon$ for all $\epsilon > 0$.
So $a_i \le -1$.
And $\{a_i\}$ is monotonically increasing.
So $\{a_i\}$ converges.
That's it.  Now, I don't have any idea what it converges to.  My proof was crude and ham-fisted.  But it was a legitimate proof.
