Prove that $\log(1+\frac{1}{n})<\frac{1}{n}$

Question

Let $n$ be any positive integer. (Yeah, I'm considering two sequences) I'd like to prove that $\log(1+\frac{1}{n})<\frac{1}{n}$, which I believe is true according to my graphing calculator. But things didn't go well when I found $$\log(1+\frac{1}{n})-\frac{1}{n}=\log(1+\frac{1}{n})-\frac{1}{n}\log 10=\log(1+\frac{1}{n})-\log 10^{\frac{1}{n}}=\log\frac{1+\frac{1}{n}}{10^{\frac{1}{n}}}.$$ This seems to add no information to our problem. What else can I do? Thank you so much.

Answer 1

Let $$ f(x)=\log\left(1+\frac{1}{x}\right)-\frac{1}{x} $$ Now, $$ \lim_{x\rightarrow\infty}f(x)=0 $$ and $$ f'(x)=\frac{1}{x^2(x+1)}>0 \text{ }\forall x>0 $$ so $f$ is an increasing function with $x$-axis as the asymptote, i.e the value of $f$ increases all the way to $0$ as $x$ increases, which has us conclude $$ f(x)<0 $$ or $$ \log\left(1+\frac{1}{n}\right)<\frac{1}{n} $$

Answer 2

One way to go about it is to prove the more general inequality $$ln(1+x) < x,$$

for all positive $x$. In other words, showing that $f(x)=ln(1+x)-x < 0$.

This follows from two facts: $f(0)=0$, and the derivative $f'(x)=\frac{1}{1+x}-1$ is negative for $x>0$.

(With a tiny bit of more work, one show that $ln(1+x) \le x$ for all $x>-1$, i.e., everywhere the function is defined, with equality iff $x=0$.)

Answer 3

See this monotone increasing sequence : $$a_n=(1+\frac1n)^n$$ The limit of this sequence is $\lim_{n\to\infty}a_n=e$.
Then, $n\log_{10}(1+\frac1n)=\log_{10}(1+\frac1n)^n\le\log_{10}e<\log_{10}10=1$.

Answer 4

Let $1+\frac 1n=e^x$, then we have

$$\begin{align}&\log\left(1+\frac 1n\right)<\frac 1n\\ \iff &x<e^x-1\\ \iff &e^x>x+1\end{align}$$

and remember that,

$$e^x = \lim_{n\to\infty}\left(1+\frac xn\right)^n≥x+1$$

by Bernoulli's inequality.