# Thread: Zeta function and bernouli numbers proof

1. ## Zeta function and bernouli numbers proof

Prove that:

$\displaystyle lim_m$$\displaystyle _\rightarrow$$\displaystyle _\infty$ $\displaystyle \mid $$\displaystyle \frac{B_{2m}}{2m}$$\displaystyle \mid$$\displaystyle =$$\displaystyle \infty$

Exploit the fact that the zeta function:

$\displaystyle \zeta(2m)$$\displaystyle <$$\displaystyle 1$

and $\displaystyle (2m)$ $\displaystyle !$ $\displaystyle >$$\displaystyle \frac{(2m)^{2m}}{e^{2m}} I have to use the formula relating the zeta function to \displaystyle B_{2m} to prove this, but i cannot show that it is equal to infinity 2. Originally Posted by ulysses123 Prove that: \displaystyle lim_m$$\displaystyle _\rightarrow$$\displaystyle _\infty \displaystyle \mid$$\displaystyle \frac{B_{2m}}{2m}$$\displaystyle \mid$$\displaystyle =$$\displaystyle \infty Exploit the fact that the zeta function: \displaystyle \zeta(2m)$$\displaystyle <$$\displaystyle 1 and \displaystyle (2m) \displaystyle ! \displaystyle >$$\displaystyle \frac{(2m)^{2m}}{e^{2m}}$

I have to use the formula relating the zeta function to $\displaystyle B_{2m}$ to prove this, but i cannot show that it is equal to infinity
$\displaystyle B_{2n} = (-1)^{n+1}\frac {2(2n)!} {(2\pi)^{2n}}\zeta(2n)$. Applying Stirling's formula we get $\displaystyle |B_{2 n}| \sim 4 \sqrt{\pi n} \left(\frac{n}{ \pi e} \right)^{2n}$.

3. i cant actually view this because it says to view images i need to have more than 10 posts, i am not sure why i cant see the solution

4. Originally Posted by ulysses123
i cant actually view this because it says to view images i need to have more than 10 posts, i am not sure why i cant see the solution
I'm pretty sure that's just my signature you can't read... What can you see?

5. [IMG]file:///C:/DOCUME%7E1/Apple/LOCALS%7E1/Temp/moz-screenshot.png[/IMG][IMG]file:///C:/DOCUME%7E1/Apple/LOCALS%7E1/Temp/moz-screenshot-1.png[/IMG]only the first bit where there is the equation for B(2n)

6. Yeah, I gave you the equation for $\displaystyle B_{2n}$ and then said what it's asymptotic too. That's enough to show what you want.