Question by Nguyễn Nhật Minh - Discuss with MathYouLike

Nguyễn Nhật Minh

04/04/2017 at 06:29

\(\dfrac{3}{4}+\dfrac{5}{36}+\dfrac{7}{144}+...+\dfrac{2n+1}{n^2\left(n+1\right)^2}< 1\)(n \(\in\)N*)

Fraction inequality

List of answers

¤« 19/02/2018 at 21:29

$\frac{3}{4} + \frac{5}{36} + \frac{7}{144} + . . . + \frac{2 n + 1}{n^{2} {(n + 1)}^{2}}$

$= 1 - \frac{1}{2^{2}} + \frac{1}{2^{2}} - \frac{1}{3^{2}} + \frac{1}{3^{2}} - \frac{1}{4^{2}} + . . . + \frac{1}{{(n - 1)}^{2}} + \frac{1}{n^{2}}$

$= 1 - \frac{1}{n^{2}} < 1$
Lê Quốc Trần Anh Coordinator 12/02/2018 at 09:27

\(\dfrac{3}{4}+\dfrac{5}{36}+\dfrac{7}{144}+...+\dfrac{2n+1}{n^2\left(n+1\right)^2}\)

\(=1-\dfrac{1}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{3^2}+\dfrac{1}{3^2}-\dfrac{1}{4^2}+...+\dfrac{1}{\left(n-1\right)^2}+\dfrac{1}{n^2}\)

\(=1-\dfrac{1}{n^2}< 1\)
FC Alan Walker 21/02/2018 at 17:38

Ta có: \(\dfrac{2n+1}{n^2\left(n+1\right)^2}=\dfrac{\left(n+1\right)^2-n^2}{n^2\left(n+1\right)^2}=\dfrac{1}{n^2}-\dfrac{1}{\left(n+1\right)^2}\)

Do đó \(\dfrac{3}{4}+\dfrac{5}{36}+\dfrac{7}{144}+...+\dfrac{2n+1}{n^2\left(n+1\right)^2}\)

\(=1-\dfrac{1}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{3^2}+\dfrac{1}{3^2}-\dfrac{1}{4^2}+...+\dfrac{1}{n^2}-\dfrac{1}{\left(n+1\right)^2}\)

\(=1-\dfrac{1}{\left(n+1\right)^2}< 1\)

Vậy \(\dfrac{3}{4}+\dfrac{5}{36}+\dfrac{7}{144}+...+\dfrac{2n+1}{n^2\left(n+1\right)^2}< 1\forall n\in\)N*

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