Prove by induction that $\sum_{i = 1}^{n} \frac{1}{\sqrt{i}} \leq 2\sqrt{n} - 1$

Question

Prove by induction that $\sum_{i = 1}^{n} \frac{1}{\sqrt{i}} \leq 2\sqrt{n} - 1$

I want to do the $n - 1 \rightarrow n$ induction step.

But I'm confused as to what my base case is. Usually if I want to do the $n \rightarrow n + 1$ induction, step, I start with $n = 1$ as my base case. But if I want to start with $n - 1$ would I be starting with 0?...But that doesn't really make sense since $i = 1$ already.

Answer 1

Take n = 1 as your base case. Note that for n = 0 the inequality does not hold:

0 is not less than or equal to $2\sqrt{0} - 1 = -1$

Induction still works because you are assuming the $n-1$ case to imply the $n$ case. You do not need to plug $1$ into $n-1$ or anything of the sort.