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

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?

• For $n=2$ we have $1 + 1/\sqrt{2} < \sqrt{2}$ but this is not true. I'm sure it should be something like $2\sqrt{k}$ on the right hand side Mar 6, 2018 at 23:02
• perhaps you mean $>\sqrt{n}$ not $<\sqrt{n}$? Mar 6, 2018 at 23:04
• @Winther you are likely right, the sum looks like a Riemann approximation to $$\int_1^{n+1} x^{-1/2}dx = 2\sqrt{n+1}...$$ Mar 6, 2018 at 23:08
• If you want to build a proof by induction I would suggest you elaborate the induction hypothesis better than you did. I mean, define for $n=k+1$ where you want to arrive. Like, $\sqrt{k+1}=\frac{k+1}{\sqrt{k+1}}$ and build your left hand side in a way to achieve that. Jan 18, 2022 at 8:30

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