1. ## convergent sequence

I need to prove that this sequence is convergent:
(an) = (1 + 1/(sqrt(2) + 1/sqrt(3) + .... + 1/sqrt(n) - 2sqrt(n)). I know that this sequence is decreasing and bounded below. Therefore, it converges. However, I'm not sure how to show it's lower bounded. It seems to be bounded below by -2. I can show its decreasing.

Thanks for any help.

2. The problem here is the sum,
$\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}}+... + \frac{1}{\sqrt{n}}$.
We have no "nice" formula for working with $s_n$.
However, in analysis we should learn how to approximate.

Here is how we do it.*

Define the following function $f(x) = \frac{1}{\sqrt{x}}$ (see below).
Now the integral:
$\int_1^n \frac{dx}{\sqrt{x}}$
Is the area below this curve.
In this particular case it is:
$\int_1^4 \frac{dx}{\sqrt{x}} = 2\sqrt{4+1} - 2$
And the area of the rectangles is (the width is 1):
$\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sq rt{3}}$
Thus, we have:
$\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sq rt{3}} \geq \int_1^4 \frac{dx}{\sqrt{x}} = 2\sqrt{4+1} - 2$
In general we have,
$\frac{1}{\sqrt{1}}+...+\frac{1}{\sqrt{n}} \geq \int_1^{n+1}\frac{dx}{\sqrt{x}} = 2\sqrt{n+1} - 2$.
Thus,
$\frac{1}{\sqrt{1}}+...+\frac{1}{\sqrt{n}} - 2\sqrt{n} \geq 2( \sqrt{n+1} -\sqrt{n}) - 2 \geq -2$.

*)The credit goes to the proof of the Integral Test for series.