Question

Let $a_1 = 1$; and $a_n = n\bigl(a_{n-1} + 1\bigr)$ for $n = 2,3,\dots$. Define

$$P_n = \prod_{k=1}^n \Bigl(1 + \frac{1}{a_k}\Bigr).$$

Then $\displaystyle\lim_{n\to\infty} P_n$ is

• A. 1+e
• B. e
• C. 1
• D. $\infty$

Answer 1

Answer: (B)

e

Answer 2

Step 1. Compute a few terms of the sequence:

$$a_1 = 1,\quad a_2 = 2(a_1+1)=4,\quad a_3 = 3(a_2+1)=15,\quad a_4 = 4\,(15+1)=64,\dots$$

Step 2. Show the closed form

$$a_n+1 = n! \sum_{k=0}^{n}\frac{1}{k!}.$$

(This can be proved by induction.)

Step 3. Then

$$1 + \frac{1}{a_k} = \frac{a_k + 1}{a_k} = \frac{k!\sum_{i=0}^k \frac{1}{i!}}{\,k!\sum_{i=0}^k \frac{1}{i!} - 1\,}.$$

So the partial product

$$P_n = \prod_{k=1}^n\Bigl(1+\frac1{a_k}\Bigr)$$

can be evaluated numerically:

$$P_1 = 2,\; P_2 = 2.5,\; P_3 \approx 2.6667,\; P_4 \approx 2.7083,\;\dots$$

It approaches the value $e\approx2.71828$.

Conclusion. $\displaystyle\lim_{n\to\infty}P_n = e.$