If a1=1 and an=n1+an−1∀n≥2 then limn→∞⁡1+1a11+1a2…………...1+1an =

=n→∞Lta1+1a1a2+1a2…………an+1an=n→∞Lta22a33……………an+1n+1a1a2………anGTann=1+an−1∀n≥2&a1=1=n→∞Ltan+1a1(n+1)!=n→∞Lt(n+1)1+an(n+1)!=n→∞Lt1n!+ann!=n→∞Lt1n!+1(n−1)!+an−1(n−1)!Proceeding in a similar manner, weget=n→∞Lt1n!+1(n−1)!+…………+12!+11!+1=e

If a1=1 and an=n1+an−1∀n≥2 then limn→∞⁡1+1a11+1a2…………...1+1an =

=n→∞Lta1+1a1a2+1a2…………an+1an=n→∞Lta22a33……………an+1n+1a1a2………anGTann=1+an−1∀n≥2&a1=1=n→∞Ltan+1a1(n+1)!=n→∞Lt(n+1)1+an(n+1)!=n→∞Lt1n!+ann!=n→∞Lt1n!+1(n−1)!+an−1(n−1)!Proceeding in a similar manner, weget=n→∞Lt1n!+1(n−1)!+…………+12!+11!+1=e

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If $a_{1} = 1 and a_{n} = n (1 + a_{n - 1}) \forall n \geq 2$ then $\lim_{n \to \infty} (1 + \frac{1}{a_{1}}) (1 + \frac{1}{a_{2}}) \dots \dots \dots \dots . . . (1 + \frac{1}{a_{n}}) =$

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Detailed Solution

$\begin{array}{l} =_{n \to \infty}^{Lt} (\frac{a_{1} + 1}{a_{1}}) (\frac{a_{2} + 1}{a_{2}}) \dots \dots \dots \dots (\frac{a_{n} + 1}{a_{n}}) \\ =_{n \to \infty}^{Lt} (\frac{a_{2}}{2}) (\frac{a_{3}}{3}) \dots \dots \dots \dots \dots (\frac{a_{n + 1}}{n + 1}) \\ a_{1} a_{2} \dots \dots \dots a_{n} \\ (GT \frac{a_{n}}{n} = 1 + a_{n - 1} \forall n \geq 2 & a_{1} = 1) \\ =_{n \to \infty}^{Lt} \frac{a_{n + 1}}{a_{1} (n + 1)!} =_{n \to \infty}^{Lt} \frac{(n + 1) (1 + a_{n})}{(n + 1)!} \\ =_{n \to \infty}^{Lt} \frac{1}{n!} + \frac{a_{n}}{n!} \\ =_{n \to \infty}^{Lt} (\frac{1}{n!} + \frac{1}{(n - 1)!} + \frac{a_{n - 1}}{(n - 1)!}) \end{array}$
Proceeding in a similar manner, weget
$\begin{array}{l} =_{n \to \infty}^{Lt} (\frac{1}{n!} + \frac{1}{(n - 1)!} + \dots \dots \dots \dots + \frac{1}{2!} + \frac{1}{1!} + 1) \\ = e \end{array}$