# Bounding an sequence

• Mar 30th 2010, 10:45 AM
Bounding an sequence
Hello all,

I have the following sequence: $\displaystyle a_n(x)=n(x^{\frac{1}{n}}-1)$ where $\displaystyle x>1$. The objective is to prove that it is bounded above and below. Also, we need to prove that it has no smallest element. I succeeded in proving that it is bounded below: (let $\displaystyle x=1+h$ with $\displaystyle h>0$ and continue from there). For the other questions, I think it would be enough if we can prove that the sequence is decreasing. The upper bound would be simply $\displaystyle a_1 = x-1$. I tried to prove this fact but i reached: $\displaystyle a_{n+1}-a_{n}=n(x^{\frac{1}{n+1}}-x^{\frac{1}{n}})+x^{\frac{1}{n+1}}-1$. But I'm stuck there. Help please.

Thanx
• Mar 30th 2010, 11:53 AM
Drexel28
Quote:

Hello all,

I have the following sequence: $\displaystyle a_n(x)=n(x^{\frac{1}{n}}-1)$ where $\displaystyle x>1$. The objective is to prove that it is bounded above and below. Also, we need to prove that it has no smallest element. I succeeded in proving that it is bounded below: (let $\displaystyle x=1+h$ with $\displaystyle h>0$ and continue from there). For the other questions, I think it would be enough if we can prove that the sequence is decreasing. The upper bound would be simply $\displaystyle a_1 = x-1$. I tried to prove this fact but i reached: $\displaystyle a_{n+1}-a_{n}=n(x^{\frac{1}{n+1}}-x^{\frac{1}{n}})+x^{\frac{1}{n+1}}-1$. But I'm stuck there. Help please.

Thanx

Let $\displaystyle z=\frac{1}{n}$ then we have $\displaystyle \frac{x^z-1}{z}$ but $\displaystyle x=1+\varepsilon$ and so $\displaystyle \frac{(1+\varepsilon)^z-1}{z}\leqslant\frac{z\cdot\varepsilon+1-1}{z}=\varepsilon$
• Mar 30th 2010, 12:00 PM
But when using Bernoulli's inequality, you have to use > and not <. In this case, we get the lower bound which is $\displaystyle \epsilon$
• Mar 30th 2010, 02:25 PM
Drexel28
Quote:

But when using Bernoulli's inequality, you have to use > and not <. In this case, we get the lower bound which is $\displaystyle \epsilon$

Not for the case when $\displaystyle 0\leqslant z\leqslant 1$.

Theorem: Let $\displaystyle 0\leqslant \delta\leqslant 1$. Then, $\displaystyle (1+x)^\delta\leqslant 1+\delta x$.

Proof: Define $\displaystyle f(x)=1+\delta x-(1+\delta)^x$. Clearly, $\displaystyle f(0)=0$ and $\displaystyle f'(x)=\delta-\delta(1+x)^{\delta-1}\geqslant\delta-\delta=0$. The conclusion follows.
• Apr 1st 2010, 03:11 AM
xalk
Quote:

But when using Bernoulli's inequality, you have to use > and not <. In this case, we get the lower bound which is $\displaystyle \epsilon$