1. ## Proof

I have seen the proof of this, but I would like to see if anyone can do it easier..

prove that

$\forall{n}>0,\text{ }\ln\bigg(1+\frac{1}{n}\bigg)>\frac{2}{2n+1}$

2. I didn't try, but maybe by induction?

3. Originally Posted by arbolis
I didn't try, but maybe by induction?
The proof I have seen involves integrals and infinite series, induction might work, but I am not entirely sure that would be easier

4. Originally Posted by Mathstud28
I have seen the proof of this, but I would like to see if anyone can do it easier..

prove that

$\forall{n}>0,\text{ }\ln\bigg(1+\frac{1}{n}\bigg)>\frac{2}{2n+1}$
Put:

$
f(x)=\ln\left(1+\frac{1}{x}\right)-\frac{2}{2x+1}
$

It is fairly easy to show that $f(x)$ is strictly decreasing for $x > 0$, and that the limit as $x \to \infty$ is $0$. Which implies that $f(x)$ has no roots for $x > 0$ and as it is greater than zero for arbitarily small $x$ that proves your result.

RonL

5. Originally Posted by CaptainBlack
Put:

$
f(x)=\ln\left(1+\frac{1}{x}\right)-\frac{2}{2x+1}
$

It is fairly easy to show that $f(x)$ is strictly decreasing for $x > 0$, and that the limit as $x \to \infty$ is $0$. Which implies that $f(x)$ has no roots for $x > 0$ and as it is greater than zero as for arbitarily small $x$ 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

$\ln\bigg(1+\frac{1}{x}\bigg)>f(x)$

then $f(x)=\frac{2}{2x+1}$

I guess I wasnt clear enough in my post, sorry

6. Originally Posted by Mathstud28
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

$\ln\bigg(1+\frac{1}{x}\bigg)>f(x)$

then $f(x)=\frac{2}{2x+1}$

I guess I wasnt clear enough in my post, sorry
Now this is not clear, what do you wish to show?.

I have shown that:

$
\ln\left(1+\frac{1}{n}\right)>\frac{2}{2n+1}
$

for all $n>0$, which was the original question.

RonL

7. Originally Posted by CaptainBlack
Now this is not clear, what do you wish to show?.

I have shown that:

$
\ln\left(1+\frac{1}{n}\right)>\frac{2}{2n+1}
$

for all $n>0$, which was the original question.

RonL
I am sorry, I will state explicitly what I want. Lets say that I was just given $\ln\bigg(1+\frac{1}{n}\bigg)$ 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 $\ln\bigg(1+\frac{1}{n}\bigg)$ and diverged by the integral test. How would one go about taking $\ln\bigg(1+\frac{1}{n}\bigg)$ 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

8. Originally Posted by Mathstud28
prove that

$\forall{n}>0,\text{ }\ln\bigg(1+\frac{1}{n}\bigg)>\frac{2}{2n+1}$
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 $f: [a,b] \rightarrow \mathbb{R}$ be such that $f'''(x)$ exists on $[a,b].$ we know that by the mean

value theorem $f(b)-f(a)=(b-a)f'(c).$ now it can be proved that:

1) if $f''(x)f'''(x) < 0, \ \forall x \in [a,b],$ then $a < c < \frac{a+b}{2}.$

2) if $f''(x)f'''(x) > 0, \ \forall x \in [a, b],$ then $\frac{a+b}{2} < c < b.$

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?

9. Originally Posted by NonCommAlg
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 $f: [a,b] \rightarrow \mathbb{R}$ be such that $f'''(x)$ exists on $[a,b].$ we know that by the mean

value theorem $f(b)-f(a)=(b-a)f'(c).$ now it can be proved that:

1) if $f''(x)f'''(x) < 0, \ \forall x \in [a,b],$ then $a < c < \frac{a+b}{2}.$

2) if $f''(x)f'''(x) > 0, \ \forall x \in [a, b],$ then $\frac{a+b}{2} < c < b.$

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 dont want to look stupid(I probably will end up doing so) but, this does not make sense to me unless it should be

$f\bigg(\frac{a+b}{2}\bigg) due to the fact of the concavity etc.

10. Originally Posted by Mathstud28
I dont want to look stupid(I probably will end up doing so) but, this does not make sense to me unless it should be

$f\bigg(\frac{a+b}{2}\bigg) due to the fact of the concavity etc.
i guess you're reffering to part 2) of the result? the result is as i mentioned, i.e. if $f''f''' > 0$ over the interval,

then $\frac{a+b}{2} < c < b.$ the beauty of the result is that it gives a better location of $c.$ the mean value theorem can

only tell us that $a < c < b.$ but under those conditions that i mentioned, we can tell if $c$ 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.