Derivative of a ratio of geometric series I am trying to prove a theorem in my paper and am stuck at this irritating thing. Please help me.
Show that $$\frac{d}{dk}\left(\frac{\sum_{x=1}^{n} x*k^x}{\sum_{x=1}^{n} k^x}\right) > 0$$ where $n > 1, k >1$
When I just calculate the ratio, I get $$\frac{n k^{(n+1)}-(n+1) k^n+1}{(k-1) (k^n-1)}$$. A simplified version of the derivative is $$\frac{1}{(k-1)^2}-\frac{n^2 k^{(n-1)}}{(k^n-1)^2}$$. There must be a simple way to show that the derivative is positive. 
Another way, I have tried to do this is by induction. Checked that it is true when $n=2$. Assuming, it holds for $N$, and show it for $N+1$. Again, I get a complicated expression there. 
 A: Note that from AM-GM, we have
$$\dfrac{1+k+k^2+\cdots+k^{n-1}}{n} \geq \left(k^{0+1+2+\cdots+(n-1)}\right)^{1/n} = k^{n(n-1)/2n} = k^{(n-1)/2}$$
Hence, for $k>1$, we have
$$\left( k-1 \right) \left(\dfrac{1+k+k^2+\cdots+k^{n-1}}{n}\right) \geq (k-1)k^{(n-1)/2}$$
which gives us
$$k^n-1 \geq n(k-1)k^{(n-1)/2} \implies (k^n-1)^2 \geq n^2 k^{n-1}(k-1)^2 \implies\dfrac1{(k-1)^2} - \dfrac{n^2k^{n-1}}{(k^n-1)^2} \geq 0$$
which is what you wanted to prove.
A: The Holder inequality may be helpful... 
A: Let $S(x)=x+x^2+\cdots +x^n$, where $n>1$. Your objective is to prove that $$f(x)=\frac{xS'(x)}{S(x)}$$ is strictly increasing. The derivative of $f(x)$ is $$f'(x)=\frac{S(x)S'(x)+xS(x)S''(x)-x(S'(x))^2}{S(x)^2}.$$
You can prove by straightforward induction that the numerator is equal to $$x^2\left(\sum_{i=3}^{n+1}{i\choose 3}x^{i-3}+\sum_{i=1}^{n-2}{n+1-i\choose 3}x^{n-2+i}\right)$$ which is of course positive for $x>0$ (not just for $x>1$). If $n=2$ the second sum is an empty sum $0$.
In the induction proof use the identity $${n-j+1\choose 3}-{n-j-1\choose 3}=(n-j-1)^2.$$
