How check if the sequence $x_n=(1+\frac{1}{n})^{\frac{1}{n}}$ is monotonically increasing or not?

Question

I want to check if the sequence $(1+\frac{1}{n})^{\frac{1}{n}}$ is monotonically increasing. I tried computing $\frac{x_{n+1}}{x_n}$ to check if the ratio is less than 1 or greater than 1, but I am unable to simplify: $$\frac{x_{n+1}}{x_n} = \frac{n^{\frac{1}{n}}\left(n+2\right)^{\frac{1}{n+1}}}{\left(n+1\right)^{\frac{2n+1}{n\left(n+1\right)}}}$$

I also tried $x_{n+1} - x_n = \left(\frac{n+2}{n+1}\right)^{\frac{1}{n+1}}-\left(\frac{n+1}{n}\right)^{\frac{1}{n}}$

Is there any other way to check the monotonic behaviour of the sequence $x_n$?

Also I would like to know if I can check $y_n=(1-\frac{1}{n})^{\frac{1}{n}}$ using similar arguments?

Answer 1

Consider $f(x)=(1+x)^x$ for $x>0$. Then $f$ is an increasing function since it is the composition of increasing functions. Another way to see it is $$f^\prime(x)=\bigg(\ln(1+x)+\frac x{1+x}\bigg)(1+x)^x >0$$ Therefore $x_n=f(\frac 1 n)$ decreases.

Use a similar technique for $y_n$.

Answer 2

The sequence decreases if and only if $\frac{\ln\left(1+\frac1n \right)}{n}$ decreases.

let $f(x) = \frac{\ln \left( 1+\frac1x \right)}{x}$ then

\begin{align} f'(x) &= \frac{x \cdot \frac{1}{1+\frac1x}\cdot \left( -\frac1{x^2} \right) - \ln \left( 1 +\frac1x \right)}{x^2} \\ &= \frac{-\frac1{x+1}-\ln\left( 1+\frac1x\right)}{x^2} \end{align}

which is negative for $x>0$.

If we consider $$g(x) = \frac{\ln \left( 1-\frac1x \right)}{x}$$

we have \begin{align} g'(x) &= \frac{x \cdot \frac{1}{1-\frac1x}\cdot \left( \frac1{x^2} \right) - \ln \left( 1 -\frac1x \right)}{x^2} \\ &= \frac{\frac1{x-1}-\ln\left( 1-\frac1x\right)}{x^2} \end{align}

which is positive for $x>1$.

Answer 3

Here is a simpler solution (which does not use any advanced calculus).

Note that \begin{alignat*}{2} &&\frac{-1}{n} & < \frac{-1}{n+1}\\ \Rightarrow &\qquad&1-\frac{1}{n} &< 1-\frac{1}{n+1}\\ \Rightarrow &\qquad&\left(1-\frac{1}{n}\right)^{\frac{1}{n}} & < \left(1-\frac{1}{n+1}\right)^{\frac{1}{n}}\label{1}\tag{1} \end{alignat*}

Now,

\begin{alignat*}{2} \text{Since,}&\qquad& \left(1-\frac{1}{n+1}\right) & < 1\\ \Rightarrow &\qquad& \left(1-\frac{1}{n+1}\right)^{\frac{1}{n}} & < 1\\ \Rightarrow &\qquad& \left(1-\frac{1}{n+1}\right)^{\frac{1}{n}}\left(1-\frac{1}{n+1}\right) & < \left(1-\frac{1}{n+1}\right)\\ \Rightarrow &&\left(1-\frac{1}{n+1}\right)^{\frac{n+1}{n}} & < \left(1-\frac{1}{n+1}\right) \end{alignat*} taking $n+1$-th root, $$\left(1-\frac{1}{n+1}\right)^{\frac{1}{n}} < \left(1-\frac{1}{n+1}\right)^{\frac{1}{n+1}}\label{2}\tag{2}$$

Thus from (\ref{1}) and (\ref{2}), we get, $$a_n=\left(1-\frac{1}{n}\right)^{\frac{1}{n}} < \left(1-\frac{1}{n+1}\right)^{\frac{1}{n}} < \left(1-\frac{1}{n+1}\right)^{\frac{1}{n+1}}=a_{n+1}$$

Thus we get $a_{n}<a_{n+1}$. Hence the sequence $(a_n)$ is monotonically increasing.

Answer 4

Considering$$x_n=\left(1+\frac{1}{n}\right)^{\frac{1}{n}}\implies \log(x_n)=\frac{1}{n}\log\left(1+\frac{1}{n}\right)$$ $$\log(x_{n+1})-\log(x_n)=\frac{1}{n+1}\log\left(1+\frac{1}{n+1}\right)-\frac{1}{n}\log\left(1+\frac{1}{n}\right)$$ Now, use Taylor expansion for large values of $n$ to get $$\log(x_{n+1})-\log(x_n)=-\frac{2}{n^3}+\frac{9}{2 n^4}+O\left(\frac{1}{n^5}\right)$$ Continue with Taylor $$\frac{x_{n+1}}{x_n}=e^{\log(x_{n+1})-\log(x_n)}=1-\frac{2}{n^3}+\frac{9}{2 n^4}+O\left(\frac{1}{n^5}\right)\implies \frac{x_{n+1}}{x_n} <1 $$

Answer 5

$f(n)= (1+1/n)^{1/n}$.

$\log (f(n))= (1/n)( \log (1+1/n))$.

$f(n)= \exp ((1/n)\log (1+1/n)).$

Let $n_2>n_1$:

1)$\log (1+1/n_2)\lt \log (1+1/n_1)$.

($\log$ is an increasing function)

2) $(1/n_2)\log (1+1/n_2) \lt$

$ (1/n_1)\log(1+1/n_1)$.

(Since $1/n_2 \lt 1/n_1$)

3) $\exp ((1/n_2)\log (1+1/n_2))\lt$

$\exp((1/n_1)\log (1+1/n_1))$.

($\exp$ is an increasing function).

Finally:

4)$f(n_2) \lt f(n_1)$ for $n_2 >n_1$.