How to prove this approximation for a logarithm?

Question

I need to prove this approximation, but I am unable to conclude $$\log \left(1+\frac{1}{n}\right) \approx \frac{1}{n}$$

Answer 1

We have $\ln(x)=\lim_{m\to\infty}m(\sqrt[m]x-1)$.

By the generalized binomial theorem,

$$\lim_{m\to\infty}m\left(\sqrt[m]{1+\frac1n}-1\right)=\\ \lim_{m\to\infty}m\left(\frac1m\frac1n-\frac{m-1}{2m^2}\frac1{n^2}+\frac{(m-1)(2m-1)}{3!m^3}\frac1{n^3}-\frac{(m-1)(2m-1)(3m-1)}{4!m^4}\frac1{n^4}\cdots\right)\\ =\frac1n-\frac1{2n^2}+\frac1{3n^3}-\frac1{4n^4}\cdots$$

Answer 2

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.

Answer 3

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 $$=\frac{1}{c_{n}}\left(n+1-n\right)=\frac{1}{c_{n}}\approx\frac{1}{n}. $$

Answer 4

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 $n \to \infty$,

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

Answer 5

By definition,

$$e^{\ln(1+1/n)}=1+\frac1n.$$

Then, using the crude approximation* $e^x\approx1+x$,

$$1+\ln\left(1+\frac1n\right)\approx1+\frac1n.$$

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*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.

Answer 6

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 $\ln(1+\frac1n)$.