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

Try this beautiful problem from TOMATO useful for ISI B.Stat Entrance based on Calculus. You may use sequential hints to solve the problem.

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

But try the problem first…

Answer: (b)$$e$$

Source

TOMATO, Problem 709

Challenges and Thrills in Pre College Mathematics

## Try with Hints

First hint

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 ……….

Final Step

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…..

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