Show $n^{1/n}$ is decreasing.

Question

This is getting to be a habit! I have done a fair amount of looking around however I cannot comment on other peoples questions yet in order to get further help when I don't understand something. My question is essentially this one Show $n^{\frac{1}{n}}$ is decreasing for $n \ge 3$.

The answer chosen as the accepted solution helps me see where to begin, but I am unable to see where to go from there. Another thing is that in my question I am asked to show the sequence is decreasing for $n\ge3$ starting with $$(1+\frac{1}{n})^n \le n$$ for all $n\ge3$. So although the linked question provides some help, it doesn't work from this point which is why I'm not sure how to proceed.

I really appreciate the help!

Just want to say a massive thank you to Jorge, ab123 and Luiz for all your help, I don't even want to imagine how frustrating that must have been for you!

Answer 1

We want to prove $n^{1/n} \geq (n+1)^{1/(n+1)}$ for $n\geq 3$.

This is equivalent to $n^{n+1}\geq (n+1)^n$

Which is equivalent to $(\frac{n+1}{n})^n\leq n$

Notice $(\frac{n+1}{n})^n=(1+\frac{1}{n})^n$.

So we have to prove $(1+\frac{1}{n})^n\leq n$ for $n\geq 3$.

Notice that $(1+\frac{1}{n})^n=\sum\limits_{i=0}^n \binom{n}{i}\frac{1}{n}^i$ by newton's theorem.

$\sum\limits_{i=0}^n \binom{n}{i}\frac{1}{n}^i=\sum\limits_{i=0}^n \frac{n!}{i!(n-i)!n^i}=\sum\limits_{i=0}^n \frac{n(n-1)\dots(n-i+1)}{i!n^i}\leq \sum\limits_{i=0}^n \frac{1}{i!}\leq 1+\sum\limits_{i=1}^n \frac{1}{2^{i-1}}\leq1+2\leq n$

Answer 2

I am not sure whether you are looking for a solution containing more powerful tools like derivatives. I try to show you this approach anyway.

Consider the function $f(x) = x^{\frac{1}{x}}, x > 0$. When $x$ is a natural number this gives exactly your sequence.

Take $\log$ of both sides: $\log(f(x)) = \log(x^{\frac{1}{x}}) = \frac{1}{x}\log(x)$ and differentiate them:

$$\frac{f'(x)}{f(x)} = -\frac{1}{x^2}\log(x)+\frac{1}{x^2}$$ hence $$f'(x) = x^{\frac{1}{x}-2}(1-\log(x)).$$ Since $x^{\frac{1}{x}-2} > 0$, this is negative when $$1-\log(x) < 0 \rightarrow x > e.$$ So $f$ is decreasing when $x > e$. If you restrict to natural numbers, then $n^{\frac{1}{n}}$ is decreasing for $n \geq 3$.

Answer 3

Let $x=y^x,$ for $x\ge 1.$ Then we can show that $$\dfrac{dy}{dx}=\dfrac{y}{x^2}(1-\ln x).$$ It is not difficult to see that this implicit function gives a maximum for $y$ at $x=e$ then decreasing so that $y=1$ is a horizontal asymptote for the graph.

Answer 4

In this case, we have to show that a sequence $ (x_n)_{n \in \mathbb N} $ given by $ x_n := n^{1/n} $ is decreasing for $ n≥3 $. To do this, we should check that

$x_{n+1} \leq x_n $ this is $ (n+1)^{1/(n+1)} \leq n^{1/n} $.

raise both members of this inequality by $ n(n+1)$ we have

$ (n+1)^n \leq n^{(n+1)}$

$ \Leftrightarrow $

$ {(n+1)^n}/{n^n} \leq n $

$ \Leftrightarrow $

$ (1 + {1/n})^n \leq n $.

if you already know how to prove this last inequality you can use these equivalences to prove that the sequence is decreasing.

another way of proving that $ \{(1 + {1/n})^n\}_{n \in \mathbb N} $ is decreasing and converges to the number $ e $:

the inequality $ {(b^{n+1} - a^{n+1})}/{(b-a)} < (n+1)b^{(n+1)} $ for all $ 0 \leq a < b $.

To prove this use the binomial expansion

$ b^{n+1} - a^{n+1} = (b-a)(b^n +ab^{(n-1)} + a^2b^{(n-2)} + ... + a^{(n-1)}b +a^n) $

Then

$ {(b^{n+1} - a^{n+1})}/{(b-a)} < b^n +bb^{(n-1)} + b^2b^{(n-2)} + ... + b^{(n-1)}b +b^n = (n+1)b^{(n+1)} $.

Now with this inequality you can get

$ b^n [b - (n+1)(b-a)] < a^{n+1}$

if we set $a = 1 + 1/(n+1)$ and $ b = 1 + 1/n $, we have $ 0 \leq a < b $ and the term in brackets reduces to $1$ and we have

$ (1 + {1/n})^n < (1 + {1/(n+1)})^{n+1} $

so we prove that the sequence is decreasing. to show that the sequence converges, it suffices to show that it is bounded. We will again use the inequality above.

Set $ a = 1 $ and $ b = 1 + 1/(2n)$. This time the term in the brackets reduces to $1/2$, and we have

$ (1 + {1/(2n)})^n < 2 $

thus $ (1 + {1/(2n)})^{2n} < 4 $

but $ \{(1 + {1/n})^n\}_{n \in \mathbb N} $ is decreasing, then

$ (1 + {1/n})^n < (1 + {1/(2n)})^{2n} < 4 $ for all $n \in \mathbb N $.

Therefore the sequence $ \{(1 + {1/n})^n\}_{n \in \mathbb N} $ is decreasing and bounded. With this we conclude that $ \{(1 + {1/n})^n\}_{n \in \mathbb N} $ is is a convergent sequence and its limit is denoted by $ e $.

This answer complements the answer given by Jorge Fernández.

Answer 5

$n^{1/n} = \exp((1/n)\log(n)).$

Since $\exp(x)$ is strictly increasing it suffices to show that

for $m >n$:

$(1/n)\log(n) >(1/m) \log(m).$

$f(x) := (1/x)\log(x), x \ge 3.$

$f'(x) = -(1/x^2)\log(x) +1/x^2 .$

$f'(x) = (1/x^2)[1-\log(x)].$

$f'(x) <0$ for $x >3. $

$f(x) $ strictly decreasing for $x>3, $

$\rightarrow$:

For $m > n:$

$n^{1/n} > m^{1/m}$, I.e. strictly decreasing.

Answer 6

I had written the first answer below, which only uses Bernoulli's Inequality, because it was close to the approach in the question, looking at $\frac{\left(1+\frac1n\right)^{n+1}}{n+1}$ instead of $\frac{\left(1+\frac1n\right)^n}n$ (they are equal). Then I saw that $\frac{(n+1)^n}{n^{n+1}}$ was decreasing, which prompted the second, simpler answer.

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First Answer

Note that $\left(1+\frac1n\right)^{n+1}$ is decreasing: $$ \begin{align} \frac{\left(1+\frac1{n-1}\right)^n}{\left(1+\frac1n\right)^{n+1}} &=\frac{n-1}n\left(1+\frac1{(n-1)(n+1)}\right)^{n+1}\tag{1a}\\ &\ge\frac{n-1}n\left(1+\frac1{n-1}\right)\tag{1b}\\[9pt] &=1\tag{1c} \end{align} $$ Explanation:
$\text{(1a):}$ rewrite the fraction
$\text{(1b):}$ Bernoulli's Inequality
$\text{(1c):}$ simplify

If $f(n)$ is decreasing, then for $n\ge1$, $f(n)\le f(1)$; that is, $$ \left(1+\frac1n\right)^{n+1}\le4\tag2 $$ Therefore, for $n\ge3$. $$ \begin{align} \frac{(n+1)^n}{n^{n+1}} &=\frac{\left(1+\frac1n\right)^{n+1}}{n+1}\tag{3a}\\ &\le\frac4{n+1}\tag{3b}\\[6pt] &\le1\tag{3c} \end{align} $$ Explanation:
$\text{(3a):}$ rewrite the fraction
$\text{(3b):}$ apply $(2)$
$\text{(3c):}$ $n\ge3$

Inequality $(3)$ says that, for $n\ge3$, $n^{1/n}$ is decreasing; that is, $$ (n+1)^{1/(n+1)}\le n^{1/n}\tag4 $$

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Second Answer

Note that $\frac{(n+1)^n}{n^{n+1}}$ is decreasing: $$ \begin{align} \frac{\frac{n^{n-1}}{(n-1)^n}}{\frac{(n+1)^n}{n^{n+1}}} &=\left(\frac{n^2}{n^2-1}\right)^n\tag{5a}\\ &\gt1\tag{5b} \end{align} $$ Thus, for $n\ge3$, $$ \begin{align} \frac{(n+1)^n}{n^{n+1}} &\le\frac{4^3}{3^4}\tag{6a}\\ &=\frac{64}{81}\tag{6b}\\[6pt] &\lt1\tag{6c} \end{align} $$ Explanation:
$\text{(6a):}$ if $f(n)$ is decreasing, then for $n\ge3$, $f(n)\le f(3)$
$\text{(6b):}$ evaluate
$\text{(6c):}$ compare

Inequality $(6)$ says that, for $n\ge3$, $n^{1/n}$ is decreasing; that is, $$ (n+1)^{1/(n+1)}\lt n^{1/n}\tag7 $$