Q1270 (Q1270): Difference between revisions

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Revision as of 10:46, 14 May 2020 Scharpf (talk | contribs) 311 edits ‎Created claim: relevant answers (P14): https://math.stackexchange.com/a/3124622 ← Older edit		Revision as of 10:49, 14 May 2020 Scharpf (talk | contribs) 311 edits ‎Created claim: relevant answers (P14): https://math.stackexchange.com/a/381303 Newer edit →
		Property / relevant answers
			https://math.stackexchange.com/a/381303
		Property / relevant answers: https://math.stackexchange.com/a/381303 / rank
			Normal rank
		Property / relevant answers: https://math.stackexchange.com/a/381303 / qualifier
			relevant answers: \lim_{n\to\infty}\log a_n=\sum_{k=1}^\infty \log \left(1-\dfrac{1}{\sqrt{k+1}}\right)

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Revision as of 10:49, 14 May 2020

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Statements

math-stackexchange-category

calculus

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sequences-and-series

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limits

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products

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topic-id

A.60

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post type

Question

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math stackexcange post id

3124615

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relevant answers

https://math.stackexchange.com/a/3124622

Defining formula

${\displaystyle 0<a_{n}=\prod _{k=1}^{n}\left(1-{\frac {1}{\sqrt {k+1}}}\right)<\prod _{k=1}^{n}\left(1-{\frac {1}{k+1}}\right)={\frac {1}{n+1}}\rightarrow 0}$

0 references

https://math.stackexchange.com/a/381303

relevant answers

\lim_{n\to\infty}\log a_n=\sum_{k=1}^\infty \log \left(1-\dfrac{1}{\sqrt{k+1}}\right)

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