Closed-Form Present Value of an Alternating Cash-Flow Series

Summary

The response derives a closed-form present value for an infinite sequence whose cash flows follow a cosine pattern and are discounted at a constant rate. It notes that odd-indexed terms vanish, rewrites the surviving cosine terms as alternating signs, and expresses the discounted payments as a geometric series. Summing that series gives a negative value for the present value when the payment scale is positive.

The derivation also identifies a convergence condition: the discount rate must be positive for the infinite series to converge as stated. This is an algebraic result for the specified periodic pattern, not a general valuation rule for arbitrary cash flows. The source provides the transformation and closed-form expression as its evidence, but no numerical example or discussion of finite payment horizons, alternative timing conventions, or cash-flow patterns. Applying the result elsewhere requires checking that the payment schedule and discounting assumptions match.

Key ideas

• Odd-indexed cosine terms vanish, leaving only every second cash flow.
• The remaining cosine values can be represented as alternating signs.
• Discounting the surviving terms produces an alternating geometric series.
• The stated infinite series converges when the discount rate is positive.

Tags

# Closed-form solution for the PV of these cash flows


# Closed-form solution for the PV of these cash flows












I've been trying to find a closed-form solution for the following equation, but without any progress. I notice that the cash flows can be separated in: C1,C3,C5... which are 0; C2,C6,C10... negative; C4,C8,C12... positive. Detailed explanation would be highly appreciated.

Teacher's given solution:

## Answer by Kevin (score 2, accepted)

https://quant.stackexchange.com/a/74122

Cool question!

In the following, I first use the fact that cosine of $\frac{\pi}{2}i$ is zero for odd numbers. I then rewrite $\cos(\pi i)$ as $(-1)^i$. Then, I add the zeroth term, apply the well-known formula for an alternating geometric series (Taylor series of $\frac{1}{1+x}$) and simplify:

\begin{align*} \sum_{i=1}^\infty \frac{\cos\left(\frac{\pi}{2}i\right)}{(1+r)^i} &= \sum_{i=1}^\infty \frac{\cos\left(\frac{\pi}{2}2i\right)}{(1+r)^{2i}} \\ &= \sum_{i=1}^\infty \frac{(-1)^i}{(1+r)^{2i}}\\ &=-1+\sum_{i=0}^\infty (-1)^i\left(\frac{1}{(1+r)^{2}}\right)^i\\ &=-1+\frac{1}{1+\frac{1}{(1+r)^{2}}} \\ &=-\frac{1}{1+(r+1)^2}<0. \end{align*}

See also here.

As you see, the series is always negative. Assuming $C>0$, the PV is always negative. Clearly, the series only converges if $r>0$.

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.