Given $(12/11)^{10}=2.3872$, prove for $n\ge 11$ that $(1+1/n)^n< 3(1-1/n)$

Question

$b^{n+1}$ - $a^{n+1}$ > $(n+1)*a^n*(b-a)$ if b > a > 0

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

let $b=1+1/n; a=1+1/(n+1)$ then

$(1+1/n)^{n+1}$ > $[(n+1)*(1/n-1/(n+1)) + 1 + 1/(n+1)]*(1+1/(n+1))^n$

$(1+1/n)^{n+1}$ > $[1+1/n + 1/(n+1)]*(1+1/(n+1))^n$

$(1+1/n)^{n}$ > $[((1+1/n + 1/(n+1))/(1+1/n)]*(1+1/(n+1))^n$

$(1+1/n)^{n}$ > $[((1+1/n + 1/(n+1))/(1+1/n)^2]*[(1+1/n)*(1+1/(n+1))^{n}]$

$(1+1/(n-1))*(1+1/n)^{n}$ > $[((1+1/(n-1))*(1+1/n + 1/(n+1)))/((1+1/n)^2*(1+1/(n+1))]*[(1+1/n)*(1+1/(n+1))^{n+1}]$

it can be proved that $[((1+1/(n-1))*(1+1/n + 1/(n+1)))/((1+1/n)^2*(1+1/(n+1))] > 1$ so

$(1+1/(n-1))*(1+1/n)^{n}$ > $(1+1/n)*(1+1/(n+1))^{n+1}$

i.e $(1+1/n)^{n}/(1-1/n)$ > $(1+1/(n+1))^{n+1}/(1-1/(n+1))$

$(1+1/n)^{n}/(1-1/n)$ is a decreasing sequence

with starting case n=11 verified less than 3, so all terms in the sequence are less than 3

Q.E.D

Answer 1

Taking logs makes the problem easier. You want to show that $$n\ln\Big(1+\dfrac{1}{n}\Big)<\ln3 + \ln\Big(1-\dfrac{1}{n}\Big).$$

Now $\ln(1+x)<x$ and from this, substituting $x=\frac{1-y}{y}$, where $y<1$, $$-\ln y=\ln\Big(\dfrac{1}{y}\Big)<\dfrac{1}{y}-1.$$ Then $$n\ln\Big(1+\dfrac{1}{n}\Big)- \ln\Big(1-\dfrac{1}{n}\Big) <n\cdot \dfrac{1}{n}+\dfrac{1}{1-\dfrac{1}{n}}-1 =\dfrac{n}{n-1}.$$ Unfortunately, this is only less than $\ln 3$ for $n=12$ and above, so you will need to do the $n=11$ case by hand, using the numerical values given.

You could also use this method to prove the result by induction: then you would need to show that $$\dfrac{\Big(1+\dfrac{1}{n+1}\Big)^{n+1}}{\Big(1+\dfrac{1}{n}\Big)^{n}}<\dfrac{1-\dfrac{1}{n+1}}{1-\dfrac{1}{n}}$$ which can be written as $$\Big(1-\dfrac{1}{n^2}\Big)\Big(1-\dfrac{1}{(n+1)^2}\Big)^{n+1}<1-\dfrac{1}{n+1}.$$ Taking logs, we require: $$\ln\Big(1-\dfrac{1}{n^2}\Big)+(n+1)\ln\Big(1-\dfrac{1}{(n+1)^2}\Big)-\ln\Big(1-\dfrac{1}{n+1}\Big)<0.$$ and using $\ln\Big(1-\dfrac{1}{n^2}\Big)<-\dfrac{1}{n^2}$, $(n+1)\ln\Big(1-\dfrac{1}{(n+1)^2}\Big)<-\dfrac{1}{n+1}$ and $-\ln\Big(1-\dfrac{1}{n+1}\Big)<\dfrac{1}{1-\dfrac{1}{n+1}}-1$ gives, for the left hand side, $-\dfrac{1}{n^2(n+1)}$ showing the required inequality.

Answer 2

For $x > 1$ let

\begin{equation}f \left(x\right) = x \ln \left(1+\frac{1}{x}\right)-\ln \left(1-\frac{1}{x}\right)\end{equation}

Its derivative is

\begin{equation}{f'} \left(x\right) = \ln \left(1+\frac{1}{x}\right)-\frac{1}{x+1}-\frac{1}{x \left(x-1\right)}\end{equation}

hence

\begin{equation}{f'} \left(x\right) \leqslant \frac{1}{x}-\frac{1}{x+1}-\frac{1}{x \left(x-1\right)} =-\frac{1}{x \left({x}^{2}-1\right)} < 0\end{equation}

Thus $ f$ decreases in $ \left(1 , \infty \right)$. For $ x \geqslant 11$ we have

\begin{equation}f \left(x\right) \leqslant f \left(11\right) = 10 \ln \left(\frac{12}{11}\right)+\ln \left(\frac{12}{10}\right) = \ln \left({{\left(\frac{12}{11}\right)}^{10}}\times{1.2}\right)\end{equation}

It follows that for such $ x \geqslant 11$,

\begin{equation}{\left(1+\frac{1}{x}\right)}^{x} \leqslant {{\left(1-\frac{1}{x}\right)}\times{2.3872}}\times{1.2} \leqslant {\left(1-\frac{1}{x}\right)}\times{3}\end{equation}