1. ## Urgent help with 2 VERY difficult problems please

For school, I need to solve the following two problems:
1.) Prove, without calculator/computer, that 2010^2008<2009^2009<2008^2010.
I believe using (n+1)^(n-1)<n^n<(n-1)^(n+1) then taking the ln of all three is the way to go.
2.) Find the limit as n goes to infinity of 1/n*sqrt[(5n)!/(3n)!] the answer for this one is NOT 0.

*Anyone that has the time and kindness PLEASE help me out. Thank you very much. And the answers do not require anything past Calculus II (integrals/limits etc)

2. Originally Posted by eid03
For school, I need to solve the following two problems:
1.) Prove, without calculator/computer, that 2010^2008<2009^2009<2008^2010.
I believe using (n+1)^(n-1)<n^n<(n-1)^(n+1) then taking the ln of all three is the way to go.
2.) Find the limit as n goes to infinity of 1/n*sqrt[(5n)!/(3n)!] the answer for this one is NOT 0.

*Anyone that has the time and kindness PLEASE help me out. Thank you very much. And the answers do not require anything past Calculus II (integrals/limits etc)
The first one is actually kind of easy. Show some actual work and we'll help (it has a clever solution).

I don't understand what the second one is? It's either infinity or zero. Th syntax is unreadable.

3. The above (LHS) is equivalent to proving that $2008\ln(2010)<2009\ln(2009)$. Recall for a second that $f\left(x+\Delta x\right)\approx \Delta xf'(x)+f(x)$ and in this case ew have that $\ln\left(x+\Delta x\right)< \frac{\Delta x}{x}+f(x)$. Taking $\Delta x=1,x=2009$ wee see that $2008\ln(2010)<\frac{2008}{2009}+2008\ln(2009)$ where the conclusion readily follows.