Find the product of a sequence

Question

How to find the product of a sequence $$\frac{2-1}{2+1}{}\frac{3-1}{3+1}...\frac{n-1}{n+1}$$

The solution is $$P_{n}=\frac{2}{n(n+1)}$$

My question is can we approximate product with integration?

Answer 1

The result holds for $n=2$,

Assume it holds for $r$, thus, $P_r = \frac{2}{r(r+1)}$

But $P_{r+1} = \frac{2-1}{2+1}\cdots\frac{r+1-1}{r+1+1}$

This implies $P_{r+1} = P_r \times \frac{r+1-1}{r+1+1}$

Plugging the value of $P_r$ above, we have:

$P_{r+1} = \frac{2}{r(r+1)}\frac{r}{r+2}$

Which is: $P_{r+1} = \frac{2}{(r+1)(r+2)}$

Answer 2

You want to approximate, I am assuming?

Take its logarithm, and the product becomes a sum of logs

$$\sum _{n=2}^k \log \left(\frac{n-1}{n+1}\right)$$

Explicitly evaluate the first 2 elements (the more elements you explicitly evaluate, better it will be)

$$\log(1/3)+\log(2/4)+\sum _{n=4}^k \log \left(\frac{n-1}{n+1}\right)$$

Use the midpoint rule to approximate the summation

$$\log(1/3)+\log(2/4)+\int_{3.5}^{k+\frac{1}{2}} \log \left(1-\frac{2} {n+1}\right) \, dn$$

$$=\log(1/3)+\log(2/4)+(k-0.5) \log (k-0.5)+(-1. k-1.5) \log (k+1.5)+4.47762$$

Exponentiate this

$$\frac16 \exp\bigg((k-0.5) \log (k-0.5)+(-1. k-1.5) \log (k+1.5)+4.47762\bigg)\\=14.67 (k-0.51)^{k-0.5} (k+1.5)^{-k-1.5}$$

A comparison for $k=20$:

Exact answer is $0.0047619$, the approximation gives $0.00472828$

If you summed more than 2 elements separately, the approximation would have been far more accurate. If you summed the first, say, 10 elements, you'd have gotten about 7 significant figures. Or if you want to do even better, you can use Euler–Mascheroni constant https://en.wikipedia.org/wiki/Euler%E2%80%93Mascheroni_constant

Answer 3

Let $$P_n=\frac{2-1}{2+1}\frac{3-1}{3+1}...\frac{n-1}{n+1}$$ If we simplify and write a few more terms in we get

$$P_n=\frac{1}{3}\frac{2}{4}\frac{3}{5}\frac{4}{6}...\frac{n-3}{n-1}\frac{n-2}{n}\frac{n-1}{n+1}$$

Now we can either see that almost everything on the top and bottom will cancel, on the top everything to the right of the $3$ will cancel with everything to the left of $n-1$ on the bottom and we are left with $$P_n=\frac{1\times 2}{n\times (n+1)}=\frac{2}{n(n+1)}$$

Alternatively we can write the top and bottom as factorials:

$$P_n=\frac{(n-1)!}{(n+1)!/2}=\frac{2(n-1)!}{(n+1)n(n-1)!}=\frac{2}{n(n+1)}$$