Show that, for all $n > 1: \frac{1}{n + 1} < \log(1 + \frac1n) < \frac1n.$

Question

I'm learning calculus, specifically derivatives and applications of MVT, and need help with the following exercice:

Show that, for all $n > 1$ $$\frac{1}{n + 1} < \log(1 + \frac1n) < \frac1n.$$

I tried to follow the below steps in order to prove the RHS inequality:

Proving that $f < g$ on $I$ from $a$ to $b$:

Step $1$. Prove that $f' < g'$ on $\operatorname{Int}(I)$.

Step $2$. Show that $f(a) \leq g(a)$ or that $f(a^+) \leq g(a^+)$

Following the above steps, let $f(x) = \log(1 + \frac1x)$ and $g(x) = \frac1x$, for all $x > 1$. One has

$$f'(x) = -\frac{1}{x^2 + x} \quad \text{and} \quad g'(x) = -\frac{1}{x^2}.$$

We note that, for every $x > 1$, $f'(x) > g'(x)$. Moreover, $f(1^+) = \log(2) < 1 = g(1^+).$

My problem is that I got the wrong inequality sign in Step $1$.

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Looking at the solution in my textbook, the author suggests using the MVT but I don't know how to apply it in this case.

Answer 1

As suggested user84413 you can apply the MVT to $f(x)=\ln x$ on the interval $[n,n+1]$ for $n>0$: there is $t\in(n,n+1)$ such that $$\log\left(1 + \frac1n\right)=\ln(n+1)-\ln(n)=f(n+1)-f(n)=f'(t)((n+1)-n)=\frac{1}{t}.$$ Now note that $0<n<t<n+1$ implies that $\frac{1}{n+1}<\frac{1}{t}<\frac{1}{n}$.

Answer 2

By definition, for any $n>0$, $$\ln {\left(1+\frac{1}{n}\right)}=\int_1^{1+\frac{1}{n}}\frac{1}{t}\,dt$$

And $$1<t<1+\frac{1}{n}\implies \frac{n}{n+1}<\frac{1}{t}<1$$

Therefore, $$\frac{n}{n+1}\int_1^{1+\frac{1}{n}}\,dt<\int_1^{1+\frac{1}{n}}\frac{1}{t}\,dt<\int_1^{1+\frac{1}{n}}\,dt$$

That is, $$\frac{1}{n+1}<\ln {\left(1+\frac{1}{n}\right)}<\frac{1}{n}$$

Answer 3

It's given in the question that $n>1$. I'm assuming $n\in\mathbb Z$.

$x_n=\left(1+\frac{1}{n}\right)^n$ is strictly increasing. Proof: $x_n>0,$ $\forall n$.

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

By Bernoulli's inequality:

$$>\frac{n+2}{n+1}\left(1-\frac{n}{(n+1)^2}\right)=$$

$$=\frac{n^3+3n^2+3n+2}{n^3+3n^2+3n+1}>1$$

By Binomial theorem: $$\left(1+\frac{1}{n}\right)^n=1+\frac{1}{n}+\frac{n(n-1)}{2!}\left(\frac{1}{n}\right)^2+$$

$$+\frac{n(n-1)(n-2)}{3!}\left(\frac{1}{n}\right)^3+\cdots+\left(\frac{1}{n}\right)^n<$$

$$<1+1+\frac{1}{2!}+\frac{1}{3!}+\cdots+\frac{1}{n!}<$$

$$<1+1+\frac{1}{2}+\frac{1}{2\cdot 3}+$$

$$+\frac{1}{3\cdot 4}+\cdots+\frac{1}{(n-1)n}=$$

Use the formula/identity $\frac{1}{(k-1)k}=\frac{1}{k-1}-\frac{1}{k}$, $\forall k$, $k\neq 0$, $k\neq 1$.

$$=1+1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+$$

$$+\frac{1}{3}-\frac{1}{4}+\cdots+\frac{1}{n-1}-\frac{1}{n}=$$

$$=3-\frac{1}{n}<3$$

Therefore $x_n$ converges. $e$ is sometimes defined as:

$$\lim_{n\to +\infty} x_n=e$$

$y_n=\left(1+\frac{1}{n}\right)^{n+1}$ is strictly decreasing. Proof: $y_n>0,$ $\forall n$.

$$\frac{y_{n-1}}{y_n}=\frac{n}{n+1}\left(1+\frac{1}{n^2-1}\right)^n>$$

By Bernoulli's inequality:

$$>\frac{n}{n+1}\left(1+\frac{n}{n^2-1}\right)=$$

$$=\frac{n^3+n^2-n}{n^3+n^2-n-1}>1$$

because $n^3+n^2-n-1>0$ because $n^3>n$, $n^2>n$, $n>1$.

$$\lim_{n\to +\infty}y_n=\lim_{n\to +\infty}\left(x_n\left(1+\frac{1}{n}\right)\right)=$$

$$=e(1+0)=e$$

$$x_n<e<y_n$$

$$\ln x_n< 1<\ln y_n$$

$$n\ln\left(1+\frac{1}{n}\right)<1<(n+1)\ln\left(1+\frac{1}{n}\right)$$

$$\frac{1}{n+1}<\ln\left(1+\frac{1}{n}\right)<\frac{1}{n}$$

Answer 4

For any $x>0$ We have $$\frac{x}{x+1} =\int_{0}^x\frac{dt}{x+1} \le \int_{0}^x\frac{dt}{t+1} =\color{red}{\ln(x+1 )}=\int_{0}^x\frac{dt}{t+1} \le \int_{0}^x\frac{dt}{1} = x $$

that is

$$\frac{x}{x+1} \le \ln(x+1 ) \le x $$

taking $x=\frac1n$ you get

$$\frac{1}{n+1} \le \ln(\frac1n+1 ) \le \frac1n $$