# Prove $\operatorname{func}(f)=\sum_{i=1}^{m}\frac{n_i}{f^2}\left(1-\frac{1}{f}\right)^{n-n_i}$ has a maximal value.

I have two questions about the function:

$\operatorname{func}(f)=\sum_{i=1}^{m}\frac{n_i}{f^2}\left(1-\frac{1}{f}\right)^{n-n_i}$

where elements are subject to the conditions:

1. $\sum_{i=1}^m n_i=n$;

2. $n_i$ and $n$ $\in \mathbb{N}$;

3. $m$ and $f$ $\in \mathbb{Z}^+$.

I). Prove a unique maximum exists with respect to $f$ exists for fixed $n_i$, $m$.

II). Also, i would like to know if the function has a upper bound solution for $f$ when $\operatorname{func}(f)$ has a maxima.

My thought:

Find the derivative of $\operatorname{func}(f)$ of $f$, then let the result equals $0$. But it seems difficult to calculate. It is my first time use stackexchange, any help would be appreciated.

• Please share your thoughts so far :) – Shaun Jul 7 '14 at 8:36
• Thanks, I'd like to share my thoughts. I have already reedited my question as you said. – Finn Jul 7 '14 at 9:08
• To clarify: do you want to prove a maximum exists with respect to $f$ exists for fixed $n_i,m$, or do you want to prove that a maximum exists with respect to $f,n_i,m$? – Jakub Konieczny Jul 7 '14 at 9:28
• I wanna prove a maximum exists with respect to $f$ exists for fixed $n_i$, $m$. Also, i would like to know if the function has a approximate analysis solution for $f$ when $\operatorname{func}(f)$ has a maxima. – Finn Jul 7 '14 at 10:17

To show that $\text{func}(f)$ has a maxima, it is enough to show this series converges.
We can write this as $$\left(1-\dfrac{1}{f}\right)^{n+1}\sum_{i = 1}^{m} \dfrac{n_i}{f^2}\left(1-\dfrac{1}{f}\right)^{-n_i-1}$$

$$=-\left(1-\dfrac{1}{f}\right)^{n+1}\sum_{i = 1}^{m} \left(1-\dfrac{1}{f}\right)^{-n_i}$$