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Question Number 147357 by Jamshidbek last updated on 20/Jul/21

$$\mathrm{If}{\mathrm{a}}_1=1\mathrm{and}\mathrm{n}⩾1{\mathrm{a}}_{\mathrm{n}+1}=\frac{1}{1+\mathrm{n}\cdot {\mathrm{a}}_{\mathrm{n}}}$$$$\mathrm{find}{\mathrm{a}}_{\mathrm{n}}=?$$

Answered by ArielVyny last updated on 20/Jul/21

$$a_1=1=\frac{1}{1+0}$$$$a_2=\frac{1}{1+a_1}=\frac{1}{1+1}(n=1)$$$$a_3=\frac{1}{1+2a_2}=\frac{1}{1+2\frac{1}{1+1}}(n=2)$$$$a_4=\frac{1}{1+3a_3}=\frac{1}{1+3\left(\frac{1}{1+2\frac{1}{1+1}}\right)}(n=3)$$$$a_5=\frac{1}{1+4a_4}=\frac{1}{1+4\left(\frac{1}{1+3\left(\frac{1}{1+2\frac{1}{1+1}}\right)}\right)}(n=4)$$$$a_n=\frac{1}{1+{na}_{n-1}}=\frac{1}{1+n(\frac{1}{1+(n-1)\left(\frac{1}{1+(n-2)\frac{1}{1+1}}\right)...\left(\frac{1}{1+(n-k)\frac{1}{1+1}}\right)}}$$$$a_n=\frac{1}{1+n\left(\frac{1}{1+\underset{k=1}{\prod ^n}\left(\frac{1}{1+(n-k)\frac{1}{1+1}}\right)}\right)}$$$$$$

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