Decreasing sequence? Given that $0 < a < b < 1$, I need to establish that the sequence $(U\;n=b^n - a^n)$ is decreasing versus $n$.
So I take the corresponding real function $(f\;x=b^x - a^x)$, then I compute its derivative like that:
$f'(x)=b^x\times (ln\:b) - a^x\times (ln\:a)$.
Normally I should prove that $f'(x) < 0$ to show that $f$ is decreasing.
Is it right? However, I fail to establish directly from the assumptions the sign of $f'$!
Am I missing something? or there is may be another way to prove that $f' < 0$.
Thanks
 A: What about rewriting $$ U(n)= b^n-a^n = b^n \left(1-\left(\frac ab \right)^n \right) $$ 
Step 1: Here $ \left(\frac ab \right)^n $ goes to zero when $n$ goest to $\infty$ because $ b\gt a$ (by definition of the problem).        
Step 2: Then the whole parenthese goes to $1$.      
Step 3: After that, the cofactor $b^n$ of the parenthese goes to zero because also $b\lt1$ (by definition of the problem).
A: The exponential and logarithm  are both increasing so $b^x>a^x$ and $ln(b)>ln(a)$, also note that $0>ln(b)> ln(a)$ (as $0<a<b<1$). It follows that  $b^x ln(b) < a^x ln(a)$, which gives us the desire conclusion.
A: The statement to be proven has to be weakened. If, e.g., $a=0.5$ and $b=0.99$, then the sequence $s_n:=b^n-a^n$ starts in increasing mode. Therefore we only can prove that the sequence $(s_n)_{n\geq1}$ is eventually decreasing. 
Assume $n>{b\over 1-b}$. Then $n>(n+1) b$ and therefore
$$n x^{n-1}>(n+1) b x^{n-1}\geq (n+1) x^n\qquad(a\leq x\leq b)\ .$$
It follows that
$$s_n=\int_a^b n x^{n-1}\ dx>\int_a^b (n+1)x^n\ dx=s_{n+1}\ .$$
