Explained: "Consider, for all n∈N:un=∫01(1−(1−tn)1n)dtDoes the series ∑n≥1un converges..."

Zeenat Horn

Answered question

2022-02-23

Consider, for all

n \in N

u_{n} = \int_{0}^{1} (1 - {(1 - t^{n})}^{\frac{1}{n}}) d t

Does the series

\sum_{n \geq 1} u_{n}

converges ?

mastifo5h

Beginner2022-02-24Added 6 answers

We have

{(1 - t^{n})}^{\frac{1}{n}} = e^{\frac{1}{n} \log (1 - t^{n})}

\geq 1 + \frac{1}{n} \log (1 - t^{n})

from which

\int_{0}^{1} {(1 - t^{n})}^{\frac{1}{n}} d t \geq 1 + \frac{1}{n} \int_{0}^{1} \log (1 - t^{n}) d t

= 1 + \frac{1}{n^{2}} \int_{0}^{1} \frac{\log (1 - t)}{t} t^{\frac{1}{n}} d t

\geq 1 + \frac{1}{n^{2}} \int_{0}^{1} \frac{\log (1 - t)}{t} d t

= 1 - \frac{π^{2}}{6 n^{2}}

Hence,

0 \leq \int_{0}^{1} (1 - {(1 - t^{n})}^{\frac{1}{n}}) d t \leq \frac{π^{2}}{6 n^{2}}

By the comparison test, the series converges.

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