For this question, I know how to do the base case and the inductive hypothesis but I'm having trouble with the inductive step. Here is what I have so far. Can anyone please help me out?
Prove by induction that for all integers $n\ge2$
$\sqrt n < \sum_{i=1}^n \frac{1}{\sqrt n }$
Base Case:
n = 2
$\sqrt 2 < 1 + \frac{1}{\sqrt 2}$
Inductive Hypothesis:
Let k be an arbitrary integer such that $k\ge2$
$\sqrt k < \sum_{i=1}^k \frac{1}{\sqrt n }$
Inductive Step:
Show true for k+1
$\sqrt{k+1} < \sum_{i=1}^{k+1} \frac{1}{\sqrt n }$
Your inequality is backwards. After fixing this, establishing the inductive step amounts to proving the following inequality: $$ \sqrt n + \frac{1}{\sqrt{n+1}}>\sqrt{n+1}. $$ You can prove it by moving the fraction to the right side, to obtain the equivalent inequality $$ \sqrt n >\sqrt{n+1}-\frac{1}{\sqrt{n+1}}. $$ You can combine the fractions on the right side using a common denominator as follows: $$ \sqrt{n+1}-\frac{1}{\sqrt{n+1}}=\frac{n+1}{\sqrt{n+1}}-\frac{1}{\sqrt{n+1}}=\frac{n}{\sqrt{n+1}}. $$ Thus the inequality that remains to be proven is $$ \sqrt n>\frac{n}{\sqrt{n+1}}, $$ which is equivalent to $$ \sqrt{n+1}>\sqrt{n} $$ after multiply both sides by $\sqrt{n+1}$ and then dividing by $\sqrt{n}$.