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# Limit Problem | ISI-B.stat | Objective Problem 694

Try this beautiful problem on Limit, useful for ISI B.Stat Entrance.

## Limit Problem | ISI B.Stat Entrance | Problem 694

Let $a_1 = 1$ and $a_n = n(a_{n-1} + 1)$ for $n = 2, 3, ….$ Define $P_n = (1 +1/a_1)(1 + 1/a_2)….(1 + 1/a_n)$. Then $\lim\limits_{x \to \infty} {P_n}$?

• (a) $1+e$
• (b) $e$
• (c) $1$
• (d) $\infty$

### Key Concepts

Calculus

Limit

Trigonometry

Answer: (b)$e$

TOMATO, Problem 709

Challenges and Thrills in Pre College Mathematics

## Try with Hints

Given that $P_n = (1 +1/a_1)(1 + 1/a_2)….(1 + 1/a_n)$

Therefore $P_n=\frac{a_1 +1}{a_1}.\frac{a_2 +1}{a_2}.\frac{a_3 +1}{a_3}.....\frac{a_n +1}{a_n}$

Now $a_n = n(a_{n-1} + 1)$

Put $n=2$, we will get $a_1+1=\frac{a_2}{2}$

$a_2+1=\frac{a_3}{3}$...................

.............................

..............................

$a_n+1=\frac{a_n}{n}$

Therefore $P_n=\frac{a_1 +1}{a_1}.\frac{a_2 +1}{a_2}.\frac{a_3 +1}{a_3}.....\frac{a_n +1}{a_n}$

$\Rightarrow {P_n}= \frac{a_2}{2a_1}.\frac{a_3}{3a_2}.\frac{a_4}{4a_3}........\frac{a_{n+1}}{(n+1).{a_n}}$

$\Rightarrow {P_n}=\frac{a_{n+1}}{{a_1}\{2.3.4...........(n+1)\}}$

$\Rightarrow {P_n}=\frac{a_{n+1}}{\{1.2.3.4...........(n+1)\}}$ (as $a_1=1$)

$\Rightarrow {P_n}=\frac{a_{n+1}}{(n+1)!}$

$\Rightarrow {P_n}=\frac{(n+1)(a_n +1)}{(n+1)!}$

$\Rightarrow {P_n}=\frac{(a_n +1)}{n!}$

$\Rightarrow {P_n}=\frac{a_n}{n!} +\frac{1}{n!}$

$\Rightarrow {P_n}=\frac{n(a_{n-1}+1)}{n!}+\frac{1}{n!}$

$\Rightarrow {P_n}=\frac{a_{n-1}+1}{(n-1)!}+\frac{1}{n!}$

$\Rightarrow {P_n}=\frac{a_{n-1}}{(n-1)!}+\frac{1}{(n-1)!}+\frac{1}{n!}$

$\Rightarrow {P_n}=\frac{a_2}{2!}+\frac{1}{2!}+\frac{1}{3!}+.......+\frac{1}{n!}$

$\Rightarrow {P_n}=\frac{2(1+1)}{2!}+\frac{1}{2!}+....+\frac{1}{n!}$

$\Rightarrow {P_n}=1+\frac{1}{1!}+.....+\frac{1}{n!}$

Can you now finish the problem ..........

Now we have to find out $\lim\limits_{x \to \infty} {P_n}$

we know that $e^x=1+\frac{x}{1!}+\frac{x^2}{2!}+..........+\infty$

So,$e^1=1+\frac{1}{1!}+\frac{1^2}{2!}+..........+\infty$

$\lim\limits_{x \to \infty} {P_n}$=$1+\frac{1}{1!}+\frac{1^2}{2!}+..........+\infty$

$\lim\limits_{x \to \infty} {P_n}$=$e$

Therefore option (b) is correct.....