Decreasing perpetuity problem A perpetuity pays 1000 immediately. The second payment is 97% of the first payment and is made at the end of the fourth year. Each subsequent payment is 97% of the previous payment and is paid four years after the previous payment. Calculate the present value of this annuity at an annual effective rate of 8%.
My attempt:
Let X denote the present value.
$x/(1+i) + 0.97x/(1+i)^5 + 0.97^2x/(1+i)^9 + ...$ = 1000
where $i = 0.08$, however I am not sure how to solve this or if this is correct
 A: Let $v = 1/(1+i)$ be the annual present value discount factor.  Then the present value is expressed as the cash flow $$PV = 1000 + 1000(0.97)v^4 + 1000(0.97)^2 v^8 + 1000(0.97)^3 v^{12} + \cdots.$$  Note that "end of the fourth year" means that four years have elapsed from the time of the first payment, for the reason that if we say "end of the first year," the payment occurs at time $t = 1$.
In actuarial notation, we would have $$PV = 1000 \ddot a_{\overline{\infty}\rceil j},$$ where $j = ((0.97)v^4)^{-1} - 1$ is the effective periodic rate corresponding to the effective periodic present value discount factor of a level payment of $1000$, adjusted for the decrease in payments, the payment frequency, and the annual rate of interest.  Since $\ddot a_{\overline{\infty}\rceil j} = 1+ \frac{1}{j}$, we immediately obtain $$PV = 1000(1 + (((0.97)(1.08)^{-4})^{-1} - 1)^{-1}) = 3484.07.$$
A: This is a perpetuity due decreasing in geometric progression and payable less frequently than interest is convertible.
The effective interest rate per period is 
$$
i=(1+0.08)^4-1=36.05\%
$$
and the growing rate is $g=-3\%$ (decreasing).
So the perpetuity due has the present value
$$
PV=1000\frac{1+i}{i-g}=3,484.07
$$
A: Since 1000 is paid immediately the present value will be equal to 1000 plus the NPV of the stream of future payments.
The dollar value of the nth future payment is $1000\cdot 0.97^n$
Each future payment is discounted by a factor of $\frac1{1.08^{4n}}$ since payments are every 4 years
So $$NPV = 1000 + \sum_{n=1}^\infty\frac{1000\cdot 0.97^n}{1.08^{4n}}=
1000 + 1000\cdot \sum_{n=1}^\infty\biggl(\frac{0.97}{1.08^4}\biggr)^n$$
$$=1000 + 1000\cdot \sum_{n=1}^\infty0.71298^n=1000\cdot \sum_{n=0}^\infty0.71298^n$$
$$=1000\cdot \frac1{1-0.71298}=3484.07$$ using $$\sum_{n=0}^\infty\alpha^n=\frac1{1-\alpha}$$
