Floor Limit

Calculus Level 3

Let $g(n) = n - \big \lfloor \frac{n}{2} \big \rfloor+ \big \lfloor \frac{n}{3} \big \rfloor - \big \lfloor \frac{n}{4} \big \rfloor + \cdots.$

Evaluate $\displaystyle \lim_{n \to \infty}\frac{g(n)}{n}.$

\log 2

\ln 2

1

2

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Tijmen Veltman
Jan 10, 2015

Assuming the limit converges, it doesn't really matter in what way we let $n$ approach infinity. If we take $n=1!,2!,3!,4!,\ldots$ , $n$ will be divisible by any natural number in the limit, meaning we can get rid of the greatest integer brackets:

$\lim_{n\to\infty} \frac{g(n)}n$

$= \lim_{n\to\infty} \frac{n-\frac{n}2+\frac{n}3-\frac{n}4+\dots}n$

$= 1-\frac12+\frac13-\frac14+\dots$

$=\boxed{\ln 2}.$

Further information on this last identity

Floor Limit

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