Proving that:∑n=199n+1−n2n+1&lt;920

As we known (2n+1)2=4n2+4n+1&gt;4n2+4n=4n(n+1), then2n+1&gt;2n(n+1), with n∈N. Hence we have∑n=199n+1−n2n+1&lt;∑n=199n+1−n2n(n+1)=12∑n=1991n−1n+1=12(1−110)=920

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High School
Calculus and Analysis
Calculus 2
Series

Rome Frazier

Answered question

2022-02-25

Proving that:

\sum_{n = 1}^{99} \frac{\sqrt{n + 1} - \sqrt{n}}{2 n + 1} < \frac{9}{20}

Answer & Explanation

Malaika Ridley

Beginner2022-02-26Added 6 answers

As we known

{(2 n + 1)}^{2} = 4 n^{2} + 4 n + 1 > 4 n^{2} + 4 n = 4 n (n + 1)

, then

2 n + 1 > 2 \sqrt{n (n + 1)}

, with

n \in N

. Hence we have

\sum_{n = 1}^{99} \frac{\sqrt{n + 1} - \sqrt{n}}{2 n + 1} < \sum_{n = 1}^{99} \frac{\sqrt{n + 1} - \sqrt{n}}{2 \sqrt{n (n + 1)}}

= \frac{1}{2} \sum_{n = 1}^{99} \frac{1}{\sqrt{n}} - \frac{1}{\sqrt{n + 1}} = \frac{1}{2} (1 - \frac{1}{10}) = \frac{9}{20}

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