For each positive integer `n` , let `y_n=1/n((n+1)(n+2)dot(n+n))^(1/n)` For `x in R` let `[x]` be the greatest integer less than or equal to `x` . If

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For each positive integer `n` , let `y_n=1/n((n+1)(n+2)dot(n+n))^(1/n)` For `x in R` let `[x]` be the greatest integer less than or equal to `x` . If

For each positive integer `n` , let `y_n=1/n((n+1)(n+2)dot(n+n))^(1/n)` For `x in R` let `[x]` be the greatest integer less than or equal to `x` . If `(lim)_(n->oo)y_n=L` , then the value of `[L]` is ______.

Monjur Alahi

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2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Correct Answer - 1
`y_(n)=[((n+1)(n+2)………(n+n))/(n^(n))]^(1//n)`
`impliesy_(n)=((n+1)/n . (n+2)/n . ………… .(n+n)/n)^(1/n)`
`:. logy_(n)=log((n+1)/n . (n+2)/n. ………… . (n+n)/n)^(1/n)`
`=1/nsum_(r=1)^(n)"log"(n+r)/n`
`=1/nsum_(r=1)^(n)log(1+r/n)`
`:. log_(nto oo) logy_(n)=lim_(n to oo) 1/n sum_(r=1)^(n)log(1+r/n)`
`:.log(lim_(nto oo) y_(n))=int_(0)^(1)log(1+x)dx`
`:.logL=int_(1)^(2)logx dx`
`=|xlogx-x|_(1)^(2)-2log2-1`
`="log"4/e`
`implies L=4/e`
`implies[L]=1`