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Gọi \(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{31}\) là S
Ta có:
\(S=1+\left(\dfrac{1}{2}+\dfrac{1}{3}\right)+\left(\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}\right)+\left(\dfrac{1}{8}+\dfrac{1}{9}+...+\dfrac{1}{15}\right)+\left(\dfrac{1}{16}+\dfrac{1}{17}+...+\dfrac{1}{31}\right)\)
\(S< 1+\left(\dfrac{1}{2}+\dfrac{1}{2}\right)+\left(\dfrac{1}{4}+\dfrac{1}{4}+\dfrac{1}{4}+\dfrac{1}{4}\right)+\left(\dfrac{1}{8}+\dfrac{1}{8}+...+\dfrac{1}{8}\right)+\left(\dfrac{1}{16}+\dfrac{1}{16}+...+\dfrac{1}{16}\right)\)
\(S< 1+1+1+1+1\)
\(S< 5\)
Vậy \(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{31}< 5\)
\(B< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{7.8}\)
\(B< \frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{8-7}{7.8}\)
\(B< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{7}-\frac{1}{8}\)
\(B< 1-\frac{1}{8}< 1\left(dpcm\right)\)
Ta có :
\(\frac{1}{3^2}< \frac{1}{2\times3};\frac{1}{4^2}< \frac{1}{3\times4};\frac{1}{5^2}< \frac{1}{4\times5};\frac{1}{6^2}< \frac{1}{5\times6};...;\frac{1}{100^2}< \frac{1}{99\times100}\)
\(\Rightarrow\) \(\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}< \frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+\frac{1}{5\times6}+...+\frac{1}{99\times100}\)
\(\Rightarrow\) \(\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}< \frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(\Rightarrow\) \(\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}< \frac{1}{2}-\frac{1}{100}\)
\(\Rightarrow\) \(\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}< \frac{49}{100}< \frac{50}{100}=\frac{1}{2}\)
\(\Rightarrow\) \(\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}< \frac{1}{2}\)
Ta có : \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2015^2}=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{2015.2015}\)
\(< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2014.2015}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2014}-\frac{1}{2015}\)
\(=1-\frac{1}{2015}=\frac{2014}{2015}< 1\)
=> A < 1 (đpcm)
Chứng tỏ rằng :
S = \(\frac{1}{1!}\)+ \(\frac{1}{2!}\)+ \(\frac{1}{3!}\)+ ...+ \(\frac{1}{2001!}\)< 3
Ta có:
1/1! = 1
1/2! = 1/1.2
1/3! = 1/2.3
1/4! < 1/3.4
1/5! < 1/4.5
.........
1/2001! < 1/2000.2001
==> S < 1 + 1/1.2 + 1/2.3 + 1/3.4 + 1/4.5 + ... + 1/2000.2001
S < 1 + 1 - 1/2 + 1/2 - 1/3 + ... + 1/2000 - 1/2001
S < 1 + 1 - 1/2001
S < 2 - 1/2001 < 2 < 3
==> S < 3
a)Ta có: \(\frac{3}{1.4}=\frac{4-1}{1.4}=1-\frac{1}{4}\)
\(\frac{3}{4.7}=\frac{7-4}{4.7}=\frac{1}{4}-\frac{1}{7}\)
... . . . .
\(\frac{3}{n\left(n+3\right)}=\frac{1}{n}-\frac{1}{n+3}\)
\(\Leftrightarrow S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{n}-\frac{1}{n+3}< 1^{\left(đpcm\right)}\)
b) Ta có: \(S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}>\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\)
\(=\frac{1}{2}-\frac{1}{10}=\frac{2}{5}\)
Suy ra \(\frac{2}{5}< S\) (1)
Ta lại có: \(S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}\)
Mà \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}=1-\frac{1}{9}=\frac{8}{9}\)
Từ đó suy ra S < 8/9
Từ (1) và (2) suy ra đpcm
2A = 2/1.2.3 + 2/2.3.4 + 2/3.4.5 + ... + 1/18.19.20
2A = 1/1.2 - 1/2.3 + 1/2.3 - 1/3.4 + 1/3.4 - 1/4.5 +...+1/18.19 - 1/19.20
2A = 1/1.2 - 1/19.20
2A = 1/2 - 1/19.20
A = (1/2 - 1/19.20) : 2
A = 1/4 - 1/(19.20.2)
MÀ 1/(19.20.2) > 0
nên A<1/4
\(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2017^2}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2016.2017}\)
\(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2017^2}< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2016}-\frac{1}{2017}\)
\(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2017^2}< 1-\frac{1}{2017}=\frac{2016}{2017}>\frac{1}{2}\)
\(\Rightarrow\)ko thể cm
Đặt \(A=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2017^2}\)
Ta có :\(\frac{1}{2^2}=\frac{1}{4}\)
\(\frac{1}{3^2}=\frac{1}{2.3}\)
.........
\(\frac{1}{2017^2}=\frac{1}{2016.2017}\)
\(\Rightarrow A< \frac{1}{4}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2016.2017}\)
\(\Rightarrow A< \frac{1}{4}-\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2016}-\frac{1}{2017}\)
\(\Rightarrow A< \frac{1}{4}-\frac{1}{2}-\frac{1}{2017}\)
\(A=\frac{-1}{4}-\frac{1}{2017}=\frac{-2021}{8068}\)
\(\Leftrightarrow A< \frac{1}{2}\) . Vì \(\frac{-2021}{8068}< \frac{1}{2}\)