"How to find limn→∞({2n}−∑k=1n1k) ?" was answered by our community

wuntsongo0cy

Answered question

2022-04-23

How to find

lim_{n \to \infty} ({2 \sqrt{n}} - \sum_{k = 1}^{n} \frac{1}{\sqrt{k}})

Tyler Velasquez

Beginner2022-04-24Added 19 answers

Use

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

, then

2 \sqrt{n} - \sum_{k = 1}^{n} \frac{1}{\sqrt{k}} = \sum_{k = 1}^{n} (2 \sqrt{k} - 2 \sqrt{k - 1} - \frac{1}{\sqrt{k}}) = \sum_{k = 1}^{n} \frac{1}{\sqrt{k}} {(\sqrt{k} - \sqrt{k - 1})}^{2}

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

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

This shows the limit does exist and

lim_{n \to \infty} (2 \sqrt{n} - \sum_{k = 1}^{n} \frac{1}{\sqrt{k}}) = \sum_{k = 1}^{\infty} \frac{1}{\sqrt{k} {(\sqrt{k} + \sqrt{k - 1})}^{2}}

The value of this sums equals

- ζ (\frac{1}{2}) \approx 1.4603545

. This value is found by other means, though:

2 \sqrt{n} - \sum_{k = 1}^{n} \frac{1}{\sqrt{k}} = 2 \sqrt{n} - (ζ (\frac{1}{2}) - ζ (\frac{1}{2}, n + 1)) \approx - ζ (\frac{1}{2}) - \frac{1}{2 \sqrt{n}} + o (\frac{1}{n})

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