Proof help: Induction ( $ 1/ \sqrt1 + 1/\sqrt2 +....+1/\sqrt{n}$ $< \sqrt{n}$ ) for $n>1$ [duplicate]

Question

Proof: Base case: For $n = 2$, we have that:

$\dfrac 12 < 2$

$ \implies \sqrt{\dfrac 12}<\sqrt2$ $ \implies \dfrac {1}{\sqrt{2}}<\sqrt2$

Assume true for $n = k$ i.e.: $ \dfrac 1{\sqrt1} + \dfrac{1}{\sqrt2} +....+\dfrac 1{\sqrt{k}} $ $< \sqrt{k} $

Then, for $n = k+1$:

$ \dfrac 1{\sqrt1} + \dfrac 1{\sqrt2} +....+\dfrac 1{\sqrt k} + \dfrac 1{\sqrt{k+1}}$ $< \sqrt{k} + \dfrac 1{\sqrt{k+1}} $

I don't know how to proceed any further and reduce $ \sqrt{k} + \dfrac{1}{\sqrt{k+1}} $ to just $ \sqrt{k+1}$

Any hints, ideas?

Answer 1

Actually,$$\frac1{\sqrt1}+\frac1{\sqrt2}+\cdots+\frac1{\sqrt n}>\sqrt n\tag1$$for each natural $n$. It is easy to see that this holds when $n=2$. Suppose that $(1)$ holds for a certain $n$. You want to deduce from this that$$\frac1{\sqrt1}+\frac1{\sqrt2}+\cdots+\frac1{\sqrt n}+\frac1{\sqrt{n+1}}>\sqrt{n+1}.$$But, from $(1)$, all you need to prove is that $\sqrt n+\frac1{\sqrt{n+1}}>\sqrt{n+1}$. This is equivalent to $\sqrt{n+1}-\sqrt n<\frac1{\sqrt{n+1}}$. But$$\sqrt{n+1}-\sqrt n=\frac1{\sqrt{n+1}+\sqrt n}<\frac1{\sqrt{n+1}}.$$

Answer 2

Hint: $$ \sqrt{k+1}-\sqrt{k}=\frac1{\sqrt{k+1}+\sqrt{k}} $$