Question

\[\text{For } p \in N\text{, } p \ge 2 \text{, show that}\]\[p (\sqrt[p]{n+1}-1) <\frac{1}{\sqrt[p]{1}}+\frac{1}{\sqrt[p]{2^{p-1}}}+...+\frac{1}{\sqrt[p]{n^{p-1}}} < p \sqrt[p]{n}\]

Answer 1

Just so you know, try using the\text{} command when you're inputting text into \(\LaTeX\). Currently trying to work out this problem...

Answer 2

The proof goes by induction. Let me call your statement \(P(p,n)\). Take \(P(2,n)\) as our base step. We will substitute to verify.\[2(\sqrt{n+1}-1)<\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\cdots + \frac{1}{\sqrt{n}} < 2\sqrt{n}\]In order to show this, we must perform a "sub-induction proof". Let us now take \(P(2,1)\) as the second-level base step.\[2(\sqrt{2}-1)<\frac{1}{\sqrt{1}}< 2\sqrt{1}\]\[2\sqrt 2 - 2 < 1 < 2\]Now, we will show \(P(2,n)\) implies \(P(2,n+1)\). I trust that you can show that \(2\sqrt{n+2} - 2\sqrt{n+1} < \frac{1}{\sqrt{n+1}} < 2\sqrt{n+1} - 2\sqrt{n}\). Adding that inequality to \(P(2,n)\) readily gives you \(P(2,n+1)\), so we therefore know \(P(2,n)\) holds for all natural numbers \(n\). This is our first-level base step.

Let me rewrite your statement as follows, for simplicity.\[p\left((n+1)^{\frac{1}{p}}-1\right) <1^{\frac{1-p}{p}}+2^{\frac{1-p}{p}}+\cdots+n^{\frac{1-p}{p}} < p n^{\frac{1}{p}}\]To complete the induction proof, we must show that \(P(p,n)\) implies \(P(p+1,n)\). Still working on that bit, but I hope this helps!

Answer 3

I concur with yakeyglee's post. :-)

Answer 4

i want to use AM-GM or MVT

Answer 5

Divide by p the three elements of your inequality.

Let
\[f(x)= \frac{x^{-\frac{p-1}{p}}}{p}\\
\int_1^{x+1} f(t) \,
dt=(x+1)^{1/p}-1 < f(1) + f(2) + \cdots +f(x)
\]
By the upper Riemann Sum ( f is decreasing)

\[\\
\int_0^{x} f(t) \,
dt=x^{1/p} >f(0)+ f(1) + f(2) + \cdots +f(x)
\]
By the Lower Riemann Sum( f decreasing and f(0)=0.
So we are done.

Answer 6

@yakeyglee,@agentx5 take a look at the proof above.

Answer 7

f(0)=0?

Answer 8

This?\[\int\limits_0^{x} f(t) \, dt=x^{1/p} > f(1) + f(2) + \cdots +f(x)\]By the Lower Riemann Sum( f decreasing)

Answer 9

Sorry, take f(0) out of the sum. The rest should be correct.

Answer 10

Good solution :)

Answer 11

@eliassaab *bows*