We have an answer to "Does the following series converge or diverge? ∑n=1∞1n+n+1"

William Montgomery

Answered question

2022-01-21

Does the following series converge or diverge?

\sum_{n = 1}^{\infty} \frac{1}{\sqrt{n} + \sqrt{n + 1}}

Hana Larsen

Beginner2022-01-22Added 17 answers

It was not entirely obvious to me that the infinite sum of differences between square roots diverges, so I did telescoping:

\sum_{n = 1}^{\infty} (\frac{1}{\sqrt{n + 1} + \sqrt{n}}) = \sum_{n = 1}^{\infty} (\frac{\sqrt{n + 1} - \sqrt{n}}{n + 1 - n})

= \sum_{n = 1}^{\infty} (\sqrt{n + 1} - \sqrt{n})

= lim_{N \to \infty} \sum_{n = 1}^{N} (\sqrt{n + 1} - \sqrt{n})

= lim_{N \to \infty} (\sqrt{2} - \sqrt{1} + \sqrt{3} - \sqrt{2} + \dots + \sqrt{N + 1} - \sqrt{N})

= lim_{N \to \infty} (\sqrt{N + 1} - \sqrt{1})

So the limit of partial sum is equal to infinity.

Kudusind

Beginner2022-01-23Added 11 answers

For

n \geq 1

, we have

\sqrt{n} + \sqrt{n + 1} \leq 2 \sqrt{n + 1} \leq 2 (n + 1) \leq 4 n

hence

\frac{1}{\sqrt{n} + \sqrt{n + 1}} \geq \frac{1}{4 n} \geq 0

and we can conclude using the fact that the harmonic series

\sum_{k = 1}^{+ \infty} \frac{1}{k}

is divergent.

RizerMix

Expert2022-01-27Added 656 answers

It is not hard to see that

\sum_{n = 1}^{\infty} \frac{1}{\sqrt{n + 1} + \sqrt{n}} = \sum_{n = 1}^{\infty} (\sqrt{n + 1} - \sqrt{n})

As you know this series is divergent.

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