Prove the following sequence is decreasing

Question

For $r>1$, prove the sequence $$X_n=\left(1+r^n\right)^{1/n}$$ is decreasing. I understand the limit is decreasing and that the limit of this sequence is $r$. I am just not sure on the algebra. My thought is to show $X_n>X_{n+1}$ by showing $X_n-X_{n+1}>0$ for all $n$. I could also use induction; however, I am not sure how that would be done.

If someone is willing to give me a push in the right direction, it would be much appreciated!

Answer 1

Here is how you show that if $x_1$, $\ldots$, $x_k >0$ and $k \ge 2$ then the function $$(0 , \infty) \ni s \mapsto (x_1^s + \cdots +x_k^s)^{\frac{1}{s}}$$ is strictly decreasing.

Let $0< s< t$. Want to show

$$ (x_1^{s} + \cdots +x_k^s)^{\frac{1}{s}}> (x_1^{t} + \cdots +x_k^t)^{\frac{1}{t}}$$

This is equivalent to:

$$\sum_i \left( \frac{x_i^s}{x_1^{s} + \cdots +x_k^s}\right)^{\frac{t}{s}} < 1$$ and you note that $\frac{t}{s} > 1$ and $$\sum_i \left( \frac{x_i^s}{x_1^{s} + \cdots +x_k^s}\right)=1$$

Answer 2

Since $\displaystyle(1+r^n)^\frac{1}{n}=\bigg(r^n\bigg[\left(\frac{1}{r}\right)^n+1\bigg]\bigg)^\frac{1}{n}=r\big(1+s^n\big)^\frac{1}{n}$ where $s=\frac{1}{r}$ satisfies $0<s<1$,

$\hspace{.3 in}$it suffices to show that $\big(1+s^n\big)^{1/n}>(1+s^{n+1})^{1/(n+1)}$ for $0<s<1$:

Since $\big(1+s^n\big)^{n+1}>(1+s^n)^n>(1+s^{n+1})^n$ since $s^n>s^{n+1}$, $\;\;\;\;\big(1+s^n\big)^{1/n}>(1+s^{n+1})^{1/(n+1)}$.

Answer 3

It is enough to show that $(1+r^n)^{n+1}>(1+r^{n+1})^n,\;\;\;$ since then $(1+r^n)^\frac{1}{n}>(1+r^{n+1})^\frac{1}{n+1}$:

$\displaystyle\big(1+r^n\big)^{n+1}-\big(1+r^{n+1}\big)^n=\sum_{k=0}^{n+1}\binom{n+1}{k}(r^n)^k-\sum_{j=0}^n\binom{n}{j}(r^{n+1})^j$

$\displaystyle=\sum_{k=1}^n\binom{n+1}{k}r^{nk}-\sum_{j=1}^{n-1}\binom{n}{j}r^{(n+1)j}=\sum_{j=0}^{n-1}\binom{n+1}{j+1}r^{n(j+1)}-\sum_{j=1}^{n-1}\binom{n}{j}r^{(n+1)j}$

$\;\;\;\;\displaystyle=(n+1)r^n+\sum_{j=1}^{n-1}\bigg[\binom{n+1}{j+1}r^{nj+n}-\binom{n}{j}r^{nj+j}\bigg]>0$

since $\binom{n+1}{j+1}=\binom{n}{j+1}+\binom{n}{j}\implies\binom{n+1}{j+1}>\binom{n}{j}$ and since $nj+n>nj+j$ for $j<n$.