I posted this question earlier, but as I don't know if a comment reply or edit will refresh this so people actually see, I'm going to post it again in hopes that someone knows what's going on. Here's the initial question:
Replacing the sequence:
$x_{n}=1+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{n}}-2\sqrt{n},\,\,\,\, n=1,2,....$
By the corresponding series, invesigate it's convergence.
Hint: Take $a_{1}=x_{1}$ and $a_{n}=x_{n}-x_{n-1}$ for $n>1$. Then $x_{n}$ is a sequence of partial sums for $\sum_{k=1}^{\infty}a_{k}$. Use an expicit formula for $a_{n}$ and use the comparison in limit test to show that the series converges.
Solution: Computing $x_n - x_{n-1}$, we get:
$x_{n}-x_{n-1} = \left(\sum_{k=1}^{n}\frac{1}{\sqrt{k}}-2\sqrt{n}\right)-\left(\sum_{k=1}^{n-1}\frac{1}{\sqrt{k}}-2\sqrt{n-1}\right)$
$a_{n} = \left(\sum_{k=1}^{n-1}\frac{1}{\sqrt{k}}+\frac{1}{\sqrt{n}}-2\sqrt{n}\right)-\left(\sum_{k=1}^{n-1}\frac{1}{\sqrt{k}}-2\sqrt{n-1}\right)$
$a_{n} = \frac{1}{\sqrt{n}}-2\sqrt{n}+2\sqrt{n-1}$
Now, the comparison test states that:
If $\lim_{n \to \infty} \frac{a_n}{b_n} = c$ for some limit $c$, then either both $a_n$ and $b_n$ converge or diverge. Trick is, I'm really struggling to find a series $b_n$ that lets me show that $a_n$ diverges (i.e. that my fraction has a finite limit, hence $a_n$ diverges).