I have seen the proof of this, but I would like to see if anyone can do it easier..
prove that
$\displaystyle \forall{n}>0,\text{ }\ln\bigg(1+\frac{1}{n}\bigg)>\frac{2}{2n+1}$
Put:
$\displaystyle
f(x)=\ln\left(1+\frac{1}{x}\right)-\frac{2}{2x+1}
$
It is fairly easy to show that $\displaystyle f(x)$ is strictly decreasing for $\displaystyle x > 0$, and that the limit as $\displaystyle x \to \infty$ is $\displaystyle 0$. Which implies that $\displaystyle f(x)$ has no roots for $\displaystyle x > 0$ and as it is greater than zero for arbitarily small $\displaystyle x$ that proves your result.
RonL
Yeah, that is a well known method of showing that an inequality is true, but can you use a similar method to find that other function.
For example this method shows the inequality, but you cannot not use it to go from
$\displaystyle \ln\bigg(1+\frac{1}{x}\bigg)>f(x)$
then $\displaystyle f(x)=\frac{2}{2x+1}$
I guess I wasnt clear enough in my post, sorry
I am sorry, I will state explicitly what I want. Lets say that I was just given $\displaystyle \ln\bigg(1+\frac{1}{n}\bigg)$ and I needed to find some function that is always smaller than it but will also diverge by the integral test. In other words your method shows how to show that the inequality is true. But in only works given the fact taht we know the inequality, say we didnt know the function that was smaller than $\displaystyle \ln\bigg(1+\frac{1}{n}\bigg)$ and diverged by the integral test. How would one go about taking $\displaystyle \ln\bigg(1+\frac{1}{n}\bigg)$ and manipulating it to find such a function.
If that was convoluted just ignore it, there is no need to prove this, its just out of curiosity
for this particular problem CaptainBlack's solution is probably the easiest, but the problem is a very special case
of the following beautiful result: let $\displaystyle f: [a,b] \rightarrow \mathbb{R}$ be such that $\displaystyle f'''(x)$ exists on $\displaystyle [a,b].$ we know that by the mean
value theorem $\displaystyle f(b)-f(a)=(b-a)f'(c).$ now it can be proved that:
1) if $\displaystyle f''(x)f'''(x) < 0, \ \forall x \in [a,b],$ then $\displaystyle a < c < \frac{a+b}{2}.$
2) if $\displaystyle f''(x)f'''(x) > 0, \ \forall x \in [a, b],$ then $\displaystyle \frac{a+b}{2} < c < b.$
the proposer, i guess, was professor Alex Lupas, a Romanian mathematician who sadly passed away few months
ago. my proof of this result is easy to follow and i can post it here, but some of you might want to think about it
a little bit first?
i guess you're reffering to part 2) of the result? the result is as i mentioned, i.e. if $\displaystyle f''f''' > 0$ over the interval,
then $\displaystyle \frac{a+b}{2} < c < b.$ the beauty of the result is that it gives a better location of $\displaystyle c.$ the mean value theorem can
only tell us that $\displaystyle a < c < b.$ but under those conditions that i mentioned, we can tell if $\displaystyle c$ is in the first half of the
interval or in the second half. probably it'd help if you see it in a couple of examples.