Question

Evaluate the limit $\lim_{n\to \infty }\sum_{i=1}^n \frac{1}{n} \cdot \lfloor \sqrt {\frac{4i}{n}} \rfloor$

score 202 · Answer 1 · 2023-05-26 17:25:43

202

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Evaluate the limit $\lim_{n\to \infty }\sum_{i=1}^n \frac{1}{n} \cdot \lfloor \sqrt {\frac{4i}{n}} \rfloor$

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score 51 · Answer 2 · 2023-06-02 02:02:57

51

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score 80 · Answer 3 · 2023-06-04 13:31:19

80

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Prove the following definite integrals as limits of sum $\int_a^b \frac{1}{x^2}dx = 1/a - 1/b$

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105

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29

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529

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score 44 · Answer 8 · 2026-03-25 23:10:39

44

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score 56 · Answer 9 · 2023-07-15 11:08:11

56

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score 34 · Answer 10 · 2023-07-19 10:45:13

34

Views

left and right Riemann sum comparison for $\int_2^\infty 1/x^2 dx$

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score 202 · Answer 11 · 2023-07-20 12:33:46

202

Views

Why can't we compute this series using Riemann's sum?

Published on 20 Jul 2023 - 12:33

score 56 · Answer 12 · 2023-07-30 20:00:31

56

Views

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score 85 · Answer 13 · 2023-08-04 14:29:32

85

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score 175 · Answer 14 · 2023-08-05 02:31:59

175

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score 118 · Answer 15 · 2023-08-27 16:46:15

118

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List Question

Evaluate the limit $\lim_{n\to \infty }\sum_{i=1}^n \frac{1}{n} \cdot \lfloor \sqrt {\frac{4i}{n}} \rfloor$

Period function/Integrability | Real Analysis

Prove the following definite integrals as limits of sum $\int_a^b \frac{1}{x^2}dx = 1/a - 1/b$

Riemann Approximation

Number of permutations for a conditional convergence series

Difference Between Riemann Integrals and Definite Integrals

Orthogonal functions and Riemann sums

Almost everywhere differentiable functions: what can we do? Integration by parts?

solution verification of $\lim_{n\to\infty}\left(\prod_{k=0}^n \frac{2n+2k}{2n+2k+1}\right)$

left and right Riemann sum comparison for $\int_2^\infty 1/x^2 dx$

Why can't we compute this series using Riemann's sum?

Does the upper bound of lower sums and the lower bound of higher sums depends on the type of intervals (open, closed or semi-open)?

Convergence of Riemann sums with partitions with equidistant points

Upper and Lower Sums for Negative Functions

Improper integral convergence implies the existence of an infinite series which its partial sum converges

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