For n in <mrow class="MJX-TeXAtom-ORD"> <mi mathvariant="double-struck">N </mrow> , conside

Kazeljkaml5n9y

Kazeljkaml5n9y

Answered question

2022-05-14

For n in N , consider the sequence { a n } defined by:
a n = 1 n k = 1 n 1 k

Answer & Explanation

Arturo Wallace

Arturo Wallace

Beginner2022-05-15Added 17 answers

Using only elementary inequalities and no (improper) integral:
For every k + 1 k = 1 k + 1 + k and 2 k < k + 1 + k < 2 k + 1 , hence
k + 1 k < 1 2 k < k k 1 .
Summing the rightmost inequality yields
a n < 2 n k = 1 n k k 1 = 2.
Likewise, summing the leftmost inequality yields
a n > 2 n k = 1 n k + 1 k = 2 n + 1 1 n > 2 2 n .
Since 2 n 0, this proves that lim n a n = 2

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