# Proof via MVT

• November 26th 2009, 07:57 AM
zhupolongjoe
Proof via MVT
For x>0, (1/x^2) - 1/(x+1)^2 >= 2/(x+1)^3 and therefore, for any positive integer n:
(1/n^2) - 1/(n+1)^2 >= 2/(n+1)^3

Prove this fact using the Mean Value Theorem.

Well, I suppose we should start by making f(x)=2/(x+1)^3
and so the MVT says that there is some c so that
f'(c)=f(t)-f(0)/(t-0) and then f'(c)*t+f(0)=f(t), but I don't know, I'm confused. This is how most of the other proofs in the text start off....
• November 26th 2009, 08:15 AM
Plato
Quote:

Originally Posted by zhupolongjoe
For x>0, (1/x^2) - 1/(x+1)^2 >= 2/(x+1)^3 and therefore, for any positive integer n:
(1/n^2) - 1/(n+1)^2 >= 2/(n+1)^3
Prove this fact using the Mean Value Theorem.
Well, I suppose we should start by making f(x)=2/(x+1)^3

No start with $f(x)=(x)^{-2}.$
Consider the interval $[n,n+1]$
• November 26th 2009, 04:32 PM
zhupolongjoe
Ok. So then we have that there is c in this interval so

f'(c)=f(t+1)-f(t)/(t+1)-t=f(t+1)-f(t)=(1/t^2+1)-(1/t^2) =1.

Now clearly 2/(x+1)^3<1 for x>=1

Is that it?
• November 26th 2009, 05:15 PM
Plato
If $f(x)=x^{-2}$ then $f'(x)=-2x^{-3}$.
By the MVT $\left( {\exists c \in (n,n + 1} \right)\left[ {f'(c) = \frac{{ - 2}}{{c^3 }} = \frac{1}{{(n + 1)^2}} - \frac{1}{n^@}} \right]$.
Simple algebra gives us $\frac{2}{{c^3 }} = \frac{1}{n^2} - \frac{1}{{(n + 1)^2}}$
and also $n < c < n + 1\, \Rightarrow \,n^3 < c^3 < \left( {n + 1} \right)^3 \, \Rightarrow \,\frac{1}{{\left( {n + 1} \right)^3 }} < \frac{1}