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The sum is the harmonic series, which can be approximated by: k = 1 ∑ n k 1 = 1 + 2 1 + 3 1 + ⋯ + n 1 = ln n + γ where γ is the Euler–Mascheroni constant. γ ≈ 0 . 5 7 7 …
Therefore, we can rewrite the question: n → ∞ lim ln n 1 k = 1 ∑ n k 1 = n → ∞ lim ln n 1 ( ln n + γ ) = n → ∞ lim ( 1 + ln n γ ) = 1 As n approaches infinity, ln n also approaches infinity, so the ln n γ approaches zero and disappears, leaving an answer of 1