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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

Check the Answer


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.....

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