Show that a function is increasing

Question

I have the following question:

Show that if j > 0, then the function f (m) = (1 + j/m)^m is an increasing function of m.

Clearly, I have to derive the function and I obtain f '(m) = (j/m + 1)^m [log(j/m + 1) - j/(m (j/m + 1))]

However, I can't see how to show that this is strictly positive, thanks for your help.

Answer 1

Hint: By the mean value theorem $$ \log(1+x)=\frac{x}{1+c} $$ for some $0<c<x$.

Additional hint: $$ \frac{x}{1+c}>\frac{x}{1+x} $$ for $c$ in this range.

Answer 2

You can do it the way you started, you just need to keep slugging. To show that $f'(m) > 0,$ it's enough to show that the stuff inside the brackets is $> 0$. So you need $$\log \left(1+\frac{j}{m} \right)-\frac{j}{m+j}>0$$ If we write $x=j/m,$ then we need $$\log \left(1+x \right) - \frac{x}{1+x} > 0, \text {for }x>0.$$ This is straightforward. The value is $0$ at $x=0$ and you can just take a derivative to see that it increases with $x.$