Max price of a share Company is planning to pay a dividend of 5\$ per share (dividend for previous year). Investor that wants to buy a shares of this company assumes that dividend will be stable (Thus will not change in future). What is the maximum price that investor can pay for one share, assuming that he will receive a right for dividend for a previous year and expected rate of return is equal to at least 9.5%?
My attempt was to use gordon's formula i.e. $\frac{D_1}{r-g}=P_0$:
$$\frac{5}{0.095}+5\approx57.63\$ $$
Is it correct?
 A: That seems correct. The $\$5$ additional dividend just add to the price so let us concentrate on the net present value of all present and future dividend, that is the geometric series
$$\$5(1+0.905+(0.905)^2+\ldots)=\frac{\$5}{1-0.905}.$$
(You want a dollar sign in your answer, not a percentage sign)
A: Your formula, $\frac{D}{r-q}$, is the value of the sum of growing dividends. This dividends are payed for infinitely periods. But in your case the dividends do not grow. Therefore $g$ is equal to $1$: $\frac{D}{r-1}$. 
$r=1+i=1.095$, where i is the interest rate.
Calculating the future value of the dividends
The first dividend has to be compounded n-1 times to get the future value of this dividend: :$Dr^{n-1}$
The second dividend has to be compounded n-2 times to get the future value of this dividend:$Dr^{n-2}$
The n-th dividend has not to be compunded to get the future value of this dividend: :D
The sum of These compunded dividends after n years is 
$S_n=D+Dr+Dr^2+\ldots Dr^{n-2}+Dr^{n-1}$
$S_n=D\left(\color{red}1+\color{blue}{r+r^2+\ldots r^{n-2}+r^{n-1}} \right) \quad (1)$
The term in the brackets is the partial sum of a geometric series.
multiplying the equation by r
$rS_n=D\left(\color{blue}{r+r^2+\ldots r^{n-2}+r^{n-1}}+\color{red}{r^n} \right) \quad (2)$
Substract 2 from 1. The blue terms are equal. They neutralize each other.
$S_n-rS_n=D(1-r^n)$
Factor out $S_n$ on the LHS
$S_n(1-r)=D(1-r^n)$
$S_n=\frac{D(1-r^n)}{1-r}$
To get the present value $S_n$ has to be divided by $r^n$
$\frac{D(1-r^n)}{r^n}\cdot \frac{1}{1-r}$
$D\left(\frac{1}{r^n}-1\right)\cdot \frac{1}{1-r}$
Now you assume that the dividends are payed for infinitely periods. Thus n goes to infinity.
$\lim_{n \to \infty} D\left(\frac{1}{r^n}-1\right)\cdot \frac{1}{1-r}$
$=D(0-1)\cdot \frac{1}{1-r}=\frac{D}{r-1}=\frac{D}{1+i-1}=\frac{D}{i}$
The present value of the infinitely times payed dividends is 
$\frac{\$ 5}{0.095}\approx \$52.63$
The investor also gets $\$5$ for the previous dividend. Thus the maximum price the investor is willing to pay is $\$52.63+\$5=57.63$ 
Yes, your answer is correct.
