Problem on increasing and decreasing function

Question

Let $a,b$ and $c$ be fixed positive real numbers. Let $u = \frac{na}{b+nc}$ for $n\ge 1$. Then

A. u increases

B. u decreases

C. u increases first and then decreases

D. None of the above statements are necessary true

I differentiated the function with respect to $n$ and got $[\frac{ab}{(b+nc)^2}] =0$

Answer 1

$u = \frac{an}{b+cn}$

By the quotient rule:

$\frac{du}{dn} = \frac{(b+cn)\cdot a - an\cdot c}{(b+cn)^{2}}$

$\frac{du}{dn} = \frac{ab}{(b+cn)^{2}}$

$a,b,c > 0 \Rightarrow \frac{du}{dn} > 0 \forall u$

Therefore the answer is $A$, i.e. $u$ increases.

Answer 2

we have $$u_n=\frac{na}{b+nc}$$ and we get $$u_{n+1}-u_n=\frac{ab}{(cn+b+c)(cn+b)}>0$$ since $$a,b,c$$ are positive.

Answer 3

If you want to figure it out quickly without using differentiation if we observe the sequence $u_{n} = \frac{na}{b + nc}$ , and write down a few terms it is in this form $\frac{a}{b + c} , \frac{2a}{b + 2c} , \frac{3a}{b + 3c} ,...$ , which can also be written in this way ... $\frac{1}{\frac{b}{a} + \frac{c}{a}} , \frac{1}{\frac{b}{2a} + \frac{c}{a}} , \frac{1}{\frac{b}{3a} + \frac{c}{a}} ,....$ and we see that the first term in the denominator decreases making the denominator decreasing as $a,b,c$ are all fixed and hence the sequence is increasing .