Let $k\in\Bbb{N}$. Prove that $0<\frac{1}{k}-\ln(1+\frac{1}{k})<\frac{1}{2k^2}$

Question

Not sure how to approach it, tried with basic algebraic manipulation but got no where. We are learning Mean Value Theorem and Taylor's Theorem so I would believe maybe we use one of those two theorems, or it may be another method. Need a hint starting it.

Answer 1

HINT

By $\frac 1k =x$ we need to show that

$$0<x-\ln(1+x)<\frac{1}{2}x^2 \iff x-\frac12x^2\le \ln(1+x)\le x \quad x\in(0,1]$$

Answer 2

You can use the Taylor series for $\log (1+x)$ to get $$\ln\left(1+\frac 1k\right)=\frac 1k-\frac 1{2k^2}+\frac 1{3k^3}-\frac1{4k^4}+\ldots$$ Now apply the alternating series theorem which says if you truncate the alternating series the error is smaller than and of the same sign as the first neglected term.

Answer 3

Further Hint: The inequality suggested by gimusi $$x-\frac{1}{2}x^2\leq \ln(1+x)\leq x\text{ for }x\geq 0$$ can be proven via Taylor's Theorem or via the inequality $$1-x\leq \frac{1}{1+x}\leq 1\text{ for }x\geq 0\,.$$

Answer 4

Hint: $\ln\left(1+\frac1k \right)=\int_k^{k+1}\frac1x dx$

Also on $[k,k+1]$ we have $\frac1{k+1} \le \frac1x \le \frac1k $, with strict inequality over most of the interval.

Then since the interval is of length $1$, we have $$ \frac1{k+1} <\int_k^{k+1}\frac1x dx <\frac1k $$ So $$ \frac1{k+1} <\ln\left( 1+\frac1k\right) <\frac1k $$ Then see if you can manipulate this inequality to what you need.

Answer 5

Let $$ x_n=\left(1+\frac 1n\right)^n $$ It's well known that $x_n$ is strictly crescent and converges to $e$. Then $$ \left(1+\frac 1k\right)^k<e\Rightarrow k\ln\left(1+\frac 1k\right) < 1\Rightarrow \frac 1k - \ln\left(1+\frac 1k\right) > 0 $$

Consider now the function $$ f(x)=\frac 1{2x^2}-\frac 1x + \ln\left(1+\frac 1x\right) $$ then $$ f'(x)=-\frac 1{x^3} +\frac 1{x^2} -\frac 1{x(x+1)}=\frac{-x-1+x^2+x-x^2}{x^3(x+1)}<0 $$ so $f$ is strictly decreasing then

$$ f(n)>\lim_{x\rightarrow +\infty}f(x)=0\Leftrightarrow \frac 1{2x^2}>\frac 1k-\ln\left(1+\frac 1k\right) $$

Answer 6

$$ \log(1+\tfrac{1}{k})=\log(k+1)-\log(k)=\int_{k}^{k+1}\frac{dx}{x}=\int_{0}^{1}\frac{dx}{x+k} \tag{1}$$

$$ \tfrac{1}{k}-\log\left(1+\tfrac{1}{k}\right)=\int_{0}^{1}\left[\frac{1}{k}-\frac{1}{x+k}\right]\,dx=\frac{1}{k}\int_{0}^{1}\frac{x}{x+k}\,dx\tag{2} $$

$$ \frac{1}{k}\int_{0}^{1}\frac{x}{x+k}\,dx \leq \frac{1}{k}\int_{0}^{1}\frac{x}{k}\,dx = \frac{1}{2k^2}.\tag{3} $$

Improving the approximation through the Cauchy-Schwarz inequality:

$$ \tfrac{1}{k}-\log\left(1+\tfrac{1}{k}\right)\approx \frac{1}{k}\sqrt{\int_{0}^{1}x\,dx\int_{0}^{1}\frac{x\,dx }{(x+k)^2}}=\tfrac{1}{k\sqrt{2}}\sqrt{\log\left(1+\tfrac{1}{k}\right)-\tfrac{1}{k+1}} $$ we get: $$ \log\left(1+\frac{1}{k}\right)\approx \frac{1+k(5+4k)-\sqrt{(1+k)(1+9k)}}{4k^2(1+k)}.\tag{4}$$ This is very accurate for large values of $k$ and acceptable for $k\approx 1$. For instance, the error of $\log(2)\approx\frac{5-\sqrt{5}}{4}$ got by setting $k=1$ is not much larger than $2\cdot 10^{-3}$.