How to prove this approximation for a logarithm? I need to prove this approximation, but I am unable to conclude
$$\log \left(1+\frac{1}{n}\right) \approx \frac{1}{n}$$
 A: By definition,
$$e^{\ln(1+1/n)}=1+\frac1n.$$
Then, using the crude approximation* $e^x\approx1+x$,
$$1+\ln\left(1+\frac1n\right)\approx1+\frac1n.$$

*By the definition of $e$ and the binomial theorem,
$$e^x=\lim_{n\to\infty}\left(1+\frac xn\right)^n=\lim_{n\to\infty}1+x+\frac{n-1}{2n}x^2+\frac{(n-1)(n-2)}{3!n^2}x^3\cdots\approx1+x.$$
Using the next term will yield
$$1+\ln\left(1+\frac1n\right)+\frac12\ln^2\left(1+\frac1n\right)=1+\frac1n,$$
which can be solved for the logarithm.
$$\left(\ln\left(1+\frac1n\right)+1\right)^2=1+\frac2n.$$
A pretty contrived method.
A: Use the inequality $e^x\ge x+1$. Substitute in $x=\ln(1+\frac1n)$ and $x=-\ln(1+\frac1n)$ to obtain an upper and lower bound for $\ln(1+\frac1n)$.
A: We have $$\log\left(1+\frac{1}{n}\right)=\log\left(n+1\right)-\log\left(n\right)=\int_{n}^{n+1}\frac{1}{t}dt
 $$ then by first mean value theorem for integration we have that exists some $c_{n}\in\left[n,n+1\right]
 $ such that $$=\frac{1}{c_{n}}\left(n+1-n\right)=\frac{1}{c_{n}}\approx\frac{1}{n}.
 $$
A: We have $\ln(x)=\lim_{m\to\infty}m(\sqrt[m]x-1)$.
By the generalized binomial theorem,
$$\lim_{m\to\infty}m\left(\sqrt[m]{1+\frac1n}-1\right)=\\
\lim_{m\to\infty}m\left(\frac1m\frac1n-\frac{m-1}{2m^2}\frac1{n^2}+\frac{(m-1)(2m-1)}{3!m^3}\frac1{n^3}-\frac{(m-1)(2m-1)(3m-1)}{4!m^4}\frac1{n^4}\cdots\right)\\
=\frac1n-\frac1{2n^2}+\frac1{3n^3}-\frac1{4n^4}\cdots$$
A: The Taylor series for the function $\log(1+x)$ about the point $x=0$ is
$$\log(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\cdots$$
so when $|x|\ll 1$ we have $\log(1+x)\approx x$. See more at the Wikipedia article on Taylor's theorem.
A: You may apply the fact that, as $x \to a$, for any differentiable function around $a$, we have

$$
\frac{f(x)-f(a)}{x-a} \to f'(a).
$$ 

Then take $f(x)=\log (1+x)$, with $f'(x)=\dfrac1{1+x}$, giving as $x \to 0$,
$$
\frac{\log(1+x)-\log(1+0)}{x-0} \to \frac1{1+0}=1
$$ $$
\frac{\log(1+x)}x \to 1,
$$ that is, as $n \to \infty$,

$$
\log\left(1+\frac1n\right) \sim \frac1n.
$$

