Krdytkyu

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.

HINT

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

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

You can use the Taylor series for $log (1+x)$ to get
$$lnleft(1+frac 1kright)=frac 1k-frac 1{2k^2}+frac 1{3k^3}-frac1{4k^4}+ldots$$
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.

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

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

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
$$ frac1{k+1} <lnleft( 1+frac1kright) <frac1k $$
Then see if you can manipulate this inequality to what you need.

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

Consider now the function
$$
f(x)=frac 1{2x^2}-frac 1x + lnleft(1+frac 1xright)
$$
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

$$
f(n)>lim_{xrightarrow +infty}f(x)=0Leftrightarrow frac 1{2x^2}>frac 1k-lnleft(1+frac 1kright)
$$

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

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

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

Improving the approximation through the Cauchy-Schwarz inequality:

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