Return to Answer

4 deleted 9 characters in body

As a matter of fact, P. L. Chebyshev knew already that for any $\epsilon > \frac{1}{5}$, there exists an $n(\epsilon) \in \mathbb{N}$ such that for all $n\geq n(\epsilon),$

$\pi((1+\epsilon)n)-\pi(n)>0.$

In [2], one can find a short report on the problem of determining explicit values for the smallest $n(\epsilon)$ explicitly once that $\epsilon$ has been fixed.

References

[1] P. L. Chebyshev. Mémoire sur les nombres premiers. Mémoires de l'Acad. Imp. Sci. de St. Pétersbourg, VII, 1850.

[2] H. Harborth & A. Kemnitz. Calculations for Bertrand's Postulate. Mathematics Magazine, 54 (1), pp. 33-34.

3 added 20 characters in body; deleted 12 characters in body

As a matter of fact, P. L. Chebyshev knew already that for any $\epsilon > \frac{1}{5}$, there exists an $n(\epsilon) \in \mathbb{N}$ such that for all $n\geq n(\epsilon),$

$\pi((1+\epsilon)n)-\pi(n)>0.$

In [2], one can find a short account of report on the problem of determining explicit values for the smallest $n(\epsilon)$ for fixed values of once that $\epsilon$.\epsilon\$ has been fixed.

References

[1] P. L. Chebyshev. Mémoire sur les nombres premiers. Mémoires de l'Acad. Imp. Sci. de St. Pétersbourg, VII, 1850.

[2] H. Harborth & A. Kemnitz. Calculations for Bertrand's Postulate. Mathematics Magazine, 54 (1), pp. 33-34.

2 added 92 characters in body

As a matter of fact, P. L. Chebyshev knew already that for any $\epsilon > \frac{1}{5}$, there exists an $n(\epsilon) \in \mathbb{N}$ such that for all $n\geq n(\epsilon),$

$\pi((1+\epsilon)n)-\pi(n)>0.$

In [2], one can find a short account of the problem of determining explicit values for the smallest $n(\epsilon)$.n(\epsilon)$for fixed values of$\epsilon\$.

References

[1] P. L. Chebyshev. Mémoire sur les nombres premiers. Mémoires de l'Acad. Imp. Sci. de St. Pétersbourg, VII, 1850.

[2] H. Harborth & A. Kemnitz. Calculations for Bertrand's Postulate. Mathematics Magazine, 54 (1), pp. 33-34.

1

As a matter of fact, P. L. Chebyshev knew already that for any $\epsilon > \frac{1}{5}$, there exists an $n(\epsilon) \in \mathbb{N}$ such that for all $n\geq n(\epsilon),$

$\pi((1+\epsilon)n)-\pi(n)>0.$

In [2], one can find a short account of the problem of determining explicit values for the smallest $n(\epsilon)$.

References

[1] P. L. Chebyshev. Mémoire sur les nombres premiers.

[2] H. Harborth & A. Kemnitz. Calculations for Bertrand's Postulate. Mathematics Magazine, 54 (1), pp. 33-34.
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 23, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.7350587844848633, "perplexity_flag": "middle"}