Let a1=1, a2=1+a1, a3=1+a1.a2, an+1=1+a1.a2.a3....an, then 1a1+1a2+1a3+.....+upto ∞ is equal to

a1=1,a2=1+a1=2 a3=1+a1.a2=3an+1=1+a1.a2.a3...an =1+(an−1)an⇒an+1=an2−an+1⇒an+1−1=an2−an⇒1an+1−1=1an2−an =1an(an−1)=1an−1−1an⇒1an+1−1=1an−1−1an ⇒1an=1an−1−1an+1−1

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Let $a_{1} = 1, a_{2} = 1 + a_{1}, a_{3} = 1 + a_{1} . a_{2}, a_{n + 1} = 1 + a_{1} . a_{2} . a_{3} .... a_{n},$ then $\frac{1}{a_{1}} + \frac{1}{a_{2}} + \frac{1}{a_{3}} + ..... +$ upto $\infty$ is equal to

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Detailed Solution

$a_{1} = 1, a_{2} = 1 + a_{1} = 2 \begin{array}{l} a_{3} = 1 + a_{1} . a_{2} = 3 \\ a_{n + 1} = 1 + a_{1} . a_{2} . a_{3} ... a_{n} \end{array} \begin{array}{l} = 1 + (a_{n} - 1) a_{n} \Rightarrow a_{n + 1} = {a_{n}}^{2} - a_{n} + 1 \\ \Rightarrow a_{n + 1} - 1 = {a_{n}}^{2} - a_{n} \Rightarrow \frac{1}{a_{n + 1} - 1} = \frac{1}{{a_{n}}^{2} - a_{n}} \end{array} \begin{array}{l} = \frac{1}{a_{n} (a_{n} - 1)} = \frac{1}{a_{n} - 1} - \frac{1}{a_{n}} \\ \Rightarrow \frac{1}{a_{n + 1} - 1} = \frac{1}{a_{n} - 1} - \frac{1}{a_{n}} \end{array} \Rightarrow \frac{1}{a_{n}} = \frac{1}{a_{n} - 1} - \frac{1}{a_{n + 1} - 1}$