Showing $\sum\limits_{i=1}^n\frac1{\sqrt i}>\sqrt n$ by induction.

Question

I need to prove this inequality by using induction: $(\frac{1}{\sqrt1} + \frac{1}{\sqrt2} + \dots + \frac{1}{\sqrt{n}} > \sqrt{n}$ for $n\geq2$.

$$\sum\limits_{i=1}^n\frac1{\sqrt i}>\sqrt n$$

I easily get by the base case for $n=2$, you get $2 > 1/4$, which is true. Then you make the assumption to be true for $(1/\sqrt1) + (1/\sqrt2) + (1/\sqrt3) +...+ (1/\sqrt n)) > \sqrt n$

And then you have to prove for $(1/\sqrt1) + (1/\sqrt2) + (1/\sqrt3) +...+ (1/\sqrt n ) + (1/\sqrt{n+1}) > \sqrt{n+1} .$

I don't know where to go from here. any help would be appreciated. Thank you.

Answer 1

For the inductive step, you need $\sqrt n+\dfrac1{\sqrt{n+1}}>\sqrt{n+1}$.

Multiply both sides by $\sqrt{n+1}$:

$$\sqrt{n(n+1)}+1>n+1.$$

Can you prove it now?

(Hint: subtract $1$ from both sides.)

Answer 2

Notice:

$$\sum_{i=1}^{n+1} \frac{1}{\sqrt{i}} = \sum_{i=1}^{n} \frac{1}{\sqrt{i}} + \frac{1}{\sqrt{n+1}} > \sqrt{n} + \frac{1}{\sqrt{n+1}}$$

So what we have to prove is:

$$\sqrt{n} + \frac{1}{\sqrt{n+1}} > \sqrt{n+1}$$

Which can be rewritten as:

$$\sqrt{n+1} - \sqrt{n} = \frac{n+1 - n}{\sqrt{n+1}+\sqrt{n}}= \frac{1}{\sqrt{n+1}+\sqrt{n}} < \frac{1}{\sqrt{n+1}}$$