A not so familiar summation

Calculus Level 5

limn(ln(1+1n)1+ln(1+2n)2+ln(1+3n)3++ln2n) \large \lim\limits_{n \to \infty} \left( \frac{ \ln ( 1 + \frac{1}{n}) }{1} + \frac{ \ln (1 + \frac{2}{n})}{2} + \frac{ \ln(1 + \frac{3}{n})}{3} + \cdots + \frac{ \ln 2}{n} \right )

If the limit above equals to AA, find 104A \lfloor 10^4 A \rfloor .


Notation: \lfloor \cdot \rfloor denotes the floor function.


Inspiration.

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