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Question Number 141409 by Willson last updated on 18/May/21

Prove that

\forall n \in N {\int_{n}}^{n + 1} lnt dt ⩽ \ln (n + \frac{1}{2})

Answered by TheSupreme last updated on 19/May/21

\int l n (t) = x l n (t) - x

\int_{n}^{n + 1} \dots = (n + 1) l n (n + 1) - n - 1 - n l n (n) + n

= n l n (n + 1) + l n (n + 1) - n l n (n) - 1

= l n {(1 + \frac{1}{n})}^{n} + l n (n + 1) - 1

f o r n > 1 l n (t) i s m o n o t o n e a n d p o s i t i v e s o \int i s m o n o t o n e

m a x (\int) = l i \underset{x}{m} l n {(1 + \frac{1}{n})}^{n} + l n (n + 1) - 1 =

= 1 + l n (n + 1) - 1 = l n (n + 1)

\int_{n}^{n + 1} l n (t) d t ⩽ l n (n + 1)