So far I have this. I have a feeling I'm getting off track with the last two steps.
We want to prove $\sum_{i=1}^{n} \frac{1}{\sqrt{i}}$ $< 2\sqrt{n}$ for $n \geq 1$
Base Case
Prove P(1): $\sum_{i=1}^{1} \frac{1}{\sqrt{1}}$ $< 2\sqrt{1}$. We get $1 < 2$.
Induction Hypothesis
$\sum_{i=1}^{n} \frac{1}{\sqrt{n}}$ $< 2\sqrt{n}$ is true for $n \geq 1$
Induction Step
Prove P(k + 1): $\sum_{i=1}^{k + 1} \frac{1}{\sqrt{k + 1}} < 2\sqrt{k + 1} $
LHS = $\sum_{i=1}^{k + 1} \frac{1}{\sqrt{k + 1}}$
= $\sum_{i=1}^{k} \frac{1}{\sqrt{k}}$ - 1 + $\frac{1}{\sqrt{k + 1}}$ + $\frac{1}{\sqrt{k + 2}}$
$< 2\sqrt{k}$ - 1 + $\frac{1}{\sqrt{k + 1}}$ + $\frac{1}{\sqrt{k + 2}}$
You should only be doing induction on $n$, not changing the index $i$ as well. We only want to show that this holds if we add the first $n$ terms of the series. For example, your induction hypothesis should read $$\sum_{i=1}^n \frac{1}{\sqrt{i}}=\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \cdots + \frac{1}{\sqrt{n}}$$ instead of $$\sum_{i=1}^n \frac{1}{\sqrt{n}} = \frac{1}{\sqrt{n}} + \frac{1}{\sqrt{n}} + \cdots + \frac{1}{\sqrt{n}}$$
Similar story for when you do the induction step, which has the additional complication of switching from $n$ to $k$. Your induction step should read: $$\sum_{i=1}^{n+1} \frac{1}{\sqrt{i}} = \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \cdots + \frac{1}{\sqrt{n}} + \frac{1}{\sqrt{n+1}}$$
How can we use our induction hypothesis to bound this?