A perpetuity costs 77.1 and makes end-of-year payments. The perpetuity pays 1 at the end of year 2, 2 at the end of year 3, ...., n at the end of year (n+1). After year (n+1), the payments remain constant at n. The annual effective interest rate is 10.5%.

Suppose the question is asking you to find 1 discounted back 2 years, 2 discounted back 3 years, 3 discounted back 4 years,..., n discounted back n+1 years, n discounted back n+2 years, n discounted back n+3 years... and setting this equal to 77.1. You have:



How do you express this sum using a closed form expression, using the variable n? $$\frac{1}{1.105^2}+\frac{2}{1.105^3}+\frac{3}{1.105^4}+...+\frac{n-1}{1.105^n}$$ Because this is not geometric, the only idea I have is you could estimate this with $$ \int_{1}^{n-1} \frac{x}{1.105^{x+1}} \,dx $$

  • $\begingroup$ The present values of the first $n$ payments form an arithmetico-geometric sequence. $\endgroup$
    – peterwhy
    Feb 4 at 23:35
  • $\begingroup$ @peterwhy thanks never seen this before, appreciate the link $\endgroup$
    – Bobas_Pett
    Feb 5 at 0:24

2 Answers 2


$\require{enclose}$Let $i = 0.105$ be the effective annual rate of interest; then $v = 1/(1+i)$ is the effective annual present value discount factor. The equation of value is

$$\begin{align} 77.1 &= 1v^2 + 2v^3 + 3v^4 + \cdots + (n-1)v^n + nv^{n+1} + nv^{n+2} + \cdots \\ &= v^2 \left(1 + 2v + 3v^2 + \cdots + (n-1)v^{n-2} + nv^{n-1}\right) + n v^{n+1} \left(v + v^2 + \cdots \right). \tag{1} \end{align}$$

The level portion of this perpetuity-immediate is simply $$nv^{n+1} (v + v^2 + \cdots ) = n \,{}_{n+1|}a_{\enclose{actuarial}{\infty}i} = nv^{n+1} a_{\enclose{actuarial}{\infty} i} = \frac{nv^{n+1}}{i}. \tag{2}$$ Expressed as an infinite geometric series, $(2)$ is equivalent to $$nv^{n+2} \sum_{k=0}^\infty v^k = nv^{n+2} \frac{1}{1-v} = \frac{nv^{n+2}}{1 - 1/(1+i)} = \frac{(1+i)nv^{n+2}}{(1+i)-1} = \frac{nv^{n+1}}{i}.$$

As for the increasing portion of the perpetuity, this amounts to a deferred increasing annuity-immediate with deferral period of one year:

$$\begin{align} v^2\left(1 + 2v + 3v^2 + \cdots + nv^{n-1}\right) &= v (I a)_{\enclose{actuarial}{n}i} \\ &= v \frac{\ddot a_{\enclose{actuarial}{n}i} - nv^n}{i} \\ &= \frac{v}{i} \left( (1+i)\frac{1 - v^n}{i} - nv^n \right). \tag{3} \end{align}$$

However, we can also use calculus to compute the sum via differentiation of a suitable finite geometric series:

$$\begin{align} v^2 \sum_{k=0}^n kv^{k-1} &= v^2 \sum_{k=0}^n \frac{d}{dv}\left[v^k\right] \\ &= v^2 \frac{d}{dv}\left[\sum_{k=0}^n v^k\right] \\ &= v^2 \frac{d}{dv} \left[ \frac{1 - v^{n+1}}{1-v} \right] \\ &= v^2 \left( \frac{1 - (1 - n + nv)v^n}{(1-v)^2} \right). \tag{4} \end{align}$$

There are other methods of evaluating the increasing sum (e.g., perturbation).

The proof of the equality of the expressions $(3)$ and $(4)$ is left as an exercise for the reader, as is the actual solution of the equation of value $(1)$ for $n$.

  • $\begingroup$ Thanks heropup! $\endgroup$
    – Bobas_Pett
    Feb 5 at 0:28

This is for anyone who was looking for a solution without using actuary notation or looking for a more first principles approach

$$\frac{1}{1.105^2}+\frac{2}{1.105^3}+\frac{3}{1.105^4}+...+\frac{n-1}{1.105^n}+\frac{n}{1.105^{n+1}}+\frac{n}{1.105^{n+2}}+\frac{n}{1.105^{n+3}}+...$$ The sum on the left is an arithmetico-geometric series, as mentioned in the comments by peterwhy. I will prove the closed form formula for n terms generally then apply it to this problem. This series has the form: $$S=a+(a+d)r+(a+2d){r^2}+(a+3d){r^3}+...+(a+(n-1)d){r^{n-1}}\tag{1}$$ Multiply both sides by r $$Sr=a+(a+d){r^2}+(a+2d){r^3}+(a+3d){r^4}+...+(a+(n-1)d){r^{n}}\tag{2}$$ Find (1)-(2) $$S(1-r)=a+dr+d{r^2}+d{r^3}+...+d{r^{n-1}}-(a+(n-1)d){r^n}$$

$$dr+d{r^2}+d{r^3}+...+d{r^{n-1}}$$ Is geometric and the closed form formula can be found using similar process as above. The closed formula is: $$d(\frac{r-r^n}{1-r})$$ $$S(1-r)=a+d(\frac{r-r^n}{1-r})-(a+(n-1)d){r^n}$$ rearranging $$S(1-r)=a+dr(\frac{1-r^n}{1-r})-(a+nd){r^n}$$ So you have $$S=\frac{a-(a+nd){r^n}}{1-r}+dr(\frac{1-r^n}{(1-r)^2})$$ $$\frac{\frac{1}{1.105^2}}{1-\frac{1}{1.105}}-\frac{(\frac{1}{1.105^2}+(n-1)\frac{1}{1.105^2})\frac{1}{1.105^{n-1}}}{1-\frac{1}{1.105}}+\frac{\frac{1}{1.105^2}\frac{1}{1.105}(1-\frac{1}{1.105^{n-1}})}{(1-\frac{1}{1.105})^2}$$ $$=\frac{1}{.105^2}-\frac{.105n+1}{{1.105^n}(.105^2)}$$

$$\frac{1}{1.105^2}+\frac{2}{1.105^3}+\frac{3}{1.105^4}+...+\frac{n-1}{1.105^n}= \frac{1}{.105^2}-\frac{.105n+1}{{1.105^n}(.105^2)}$$ $$\frac{n}{1.105^{n+1}}+\frac{n}{1.105^{n+2}}+\frac{n}{1.105^{n+3}}+...$$ Is an infinite geometric series and is equal to $$\frac{n}{{1.105^n}(.105)}$$ So $$\frac{1}{1.105^2}+\frac{2}{1.105^3}+\frac{3}{1.105^4}+...+\frac{n-1}{1.105^n}+\frac{n}{1.105^{n+1}}+\frac{n}{1.105^{n+2}}+\frac{n}{1.105^{n+3}}+...= \frac{1}{.105^2}-\frac{.105n+1}{{1.105^n}(.105^2)}+ \frac{n}{{1.105^n}(.105)}$$ $$=\frac{1}{.105^2}-\frac{1}{{1.105^n}(.105^2)}=77.1$$ Solving for n gives $\approx19$


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