Let $k\in\Bbb{N}$. Prove that $0<\frac{1}{k}-\ln(1+\frac{1}{k})<\frac{1}{2k^2}$ 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.
 A: HINT
By $\frac 1k =x$ we need to show that
$$0<x-\ln(1+x)<\frac{1}{2}x^2 \iff x-\frac12x^2\le \ln(1+x)\le x \quad x\in(0,1]$$
A: 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.
A: 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
$$1-x\leq \frac{1}{1+x}\leq 1\text{ for }x\geq 0\,.$$
A: 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.
A: 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
$$
f(n)>\lim_{x\rightarrow +\infty}f(x)=0\Leftrightarrow \frac 1{2x^2}>\frac 1k-\ln\left(1+\frac 1k\right)
$$
A: $$ \log(1+\tfrac{1}{k})=\log(k+1)-\log(k)=\int_{k}^{k+1}\frac{dx}{x}=\int_{0}^{1}\frac{dx}{x+k} \tag{1}$$
$$ \tfrac{1}{k}-\log\left(1+\tfrac{1}{k}\right)=\int_{0}^{1}\left[\frac{1}{k}-\frac{1}{x+k}\right]\,dx=\frac{1}{k}\int_{0}^{1}\frac{x}{x+k}\,dx\tag{2} $$
$$ \frac{1}{k}\int_{0}^{1}\frac{x}{x+k}\,dx \leq \frac{1}{k}\int_{0}^{1}\frac{x}{k}\,dx = \frac{1}{2k^2}.\tag{3} $$
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 $2\cdot 10^{-3}$.
