Calculating Annuity Payments 
  
*
  
*A man turns $40$ today and wishes to provide supplemental retirement income of $3000$ at the beginning of each month starting on his $65$-th birthday. 
  
*Starting today, he makes monthly contributions of $X$ to a fund for $25$ years.
  
*The fund earns an annual nominal interest rate of $8 \%$ compounded monthly.  
  
*On his $65$-th birthday, each $1000$ of the fund will provide $9.65$ of income at the beginning of each month starting immediately and continuing as long as he survives.
  
*Calculate $X$.  
  

Hi, does any one understand the last statement regarding how each $\$\ 1000$ funds $\$\ 9.65$ of the income ?. The expression is going to look something like this:
\begin{align}
\sum_{k = 1}^{300} X\,\left(1 + {.08 \over12}\right)^{k}\ =\  ???
\end{align}
Now, the other half is supposed to be the present value of the monthly incomes.  Is the last statement saying that the annuity can sustain itself forever ?.  If that's not the case, then the payments must end at some point
( his death ), but I don't see how we can determine the duration.
 A: The cash flow of payments into the fund is finite.  The cash flow of income paid by the fund is not; it is a perpetuity because it must be sustaining for as long as he survives.
In actuarial notation, we have $$X \ddot s_{\overline{300}\rceil j} = 3000 \ddot a_{\overline{\infty}\rceil k},$$ where $j = i^{(12)}/12 = 0.08/12$, and $k = 9.65/1000 = 0.00965$.  This equation of value is written with respect to the time of the first receipt of income from the fund.  Since $$\ddot s_{\overline{n}\rceil j} = (1+j) \frac{(1+j)^n - 1}{j}, \\ \ddot a_{\overline{\infty}\rceil k} = 1 + \frac{1}{k},$$ it follows that $$X = 3000 \cdot \frac{104.627}{957.367} = 327.858\ldots.$$
A: By GP sum,
$$X(1+\frac{8\%}{12})\frac{1-(1+\frac{8\%}{12})^{300}}{1-(1+\frac{8\%}{12})}=\frac{3000}{1-(1+\frac{9.65}{1000})^{-1}}$$
then
$X=327.8586$
However, if the question is amended to "...supplemental retirement income of 3000 at the END of each month starting on his 65-th birthday...", then 324.70 would be correct.
i.e.
$$X(1+\frac{8\%}{12})\frac{1-(1+\frac{8\%}{12})^{300}}{1-(1+\frac{8\%}{12})}=\frac{\frac{3000}{1+\frac{9.65}{1000}}}{1-(1+\frac{9.65}{1000})^{-1}}$$
or
$X=327.8586(1+\frac{9.65}{1000})^{-1}=324.725$
