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\(A< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{19\cdot20}\)
\(\Rightarrow A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{20}\)
\(\Rightarrow A< 1-\frac{1}{20}< 1\)
Ta có \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{7\cdot8}\)
\(\Rightarrow A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-....+\frac{1}{7}-\frac{1}{8}\)
\(\Rightarrow A< 1-\frac{1}{8}< 1\)
Bài 2:
a, S = 1/11 + 1/12 + .. +1/20 với 1/2
SỐ số hạng tổng S: [20 - 11]: 1 + 1 = 10 số
mà 1/11 > 1/20
1/12 > 1/20
.........................
1/20 = 1/20
=> 1/11 + 1/12 + ... + 1/20 > 1/20 . 10 => S > 1/2
b, B = 2015/2016 + 2016/2017 và C = 2015+2016/2016+2017
Dễ dàng ta thấy: C = 4031/4033 < 1
B = 2015/2016 + 2016/2017
B = 2015/2016 + [1/2016 + 4062239/4066272]
B = [2015/2016 + 1/2016] + 4062239/4066272]
B = 1 +4062239/4066272
=> B > 1
Vậy B > C
c, [-1/5]^9 và [-1/25]^5
ta có: 255 = [52]5 = 52.5 = 510 > 59
=> [1/5]9 > [1/25]5
=> [-1/5]9 < [-1/25]5
d, 1/32+1/42+1/52+1/62 và 1/2
ta có: 1/3^2 + 1/4^2 + 1/5^2 + 1/6^2 = 1/9 + 1/16 + 1/25 + 1/36
mà: 1/9 < 1/8
1/16 < 1/8
1/25 < 1/8
1/36 < 1/8
=> 1/9+1/16+1/25+1/36 < 1/2
Vậy 1/32+1/42+1/52+1/62 < 1/2
Bài 1:
A = 3/4 . 8/9 . 15/16....2499/2500
A = [1.3/22][2.4/32]....[49.51/502]
A = [1.2.3.4.5...51 / 2.3.4....50][3.4.5...51 / 2.3.4...50]
A = 1/50 . 51/2
A = 51/100
B = 22/1.3 + 32/2.4 + ... + 502/49.51
B = 4/3.9/8....2500/2499
Nhận thấy B ngược A => B = 100/51 [cách tính tương tự tính A]
Bài 2:
a. S = 1/11+1/12+...+1/20 và 1/2
Số số hạng tổng S: [20 - 11]: 1 + 1 = 10 [ps]
ta có: 1/11 > 1/20
Ta có:
\(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{2017}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2018}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2018}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}\right)-\left(1+\frac{1}{2}+...+\frac{1}{1009}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}-1-\frac{1}{2}-...-\frac{1}{1009}\)
\(=\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2018}=B\)
Ta có\(\frac{1}{5}+\frac{1}{6}+...+\frac{1}{10}\)<\(\frac{1}{5}+\frac{1}{5}+\frac{1}{5}+...+\frac{1}{5}\)=\(\frac{5}{6}\)(6 c/s \(\frac{1}{5}\))
Ta lại có \(\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{17}\)<\(\frac{1}{11}+\frac{1}{11}+...+\frac{1}{11}\)=\(\frac{7}{11}\)(7 c/s \(\frac{1}{11}\))
Suy ra \(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+...+\frac{1}{17}\)<\(\frac{110}{55}\)=2
Vậy...
Hok tốt
Đặt \(A=\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+...+\frac{1}{17}\)
Ta có: \(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}< \frac{1}{5}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}=\frac{6}{5}\)
\(\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{16}+\frac{1}{17}< \frac{1}{11}+\frac{1}{11}+\frac{1}{11}+\frac{1}{11}+\frac{1}{11}+\frac{1}{11}+\frac{1}{11}=\frac{7}{11}\)
\(\Rightarrow\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+...+\frac{1}{17}< \frac{6}{5}+\frac{7}{11}\)
\(\Rightarrow A< \frac{101}{55}< \frac{110}{55}=2\)
\(\Rightarrow A< 2\)( ĐPCM )
sửa đề : \(\frac{9}{10!}+\frac{10}{11!}+\frac{11}{12!}+...+\frac{99}{100!}\)
\(=\frac{10-1}{10!}+\frac{11-1}{11!}+\frac{12-1}{12!}+...+\frac{100-1}{100!}\)
\(=\frac{1}{9!}-\frac{1}{10!}+\frac{1}{10!}-\frac{1}{11!}+\frac{1}{11!}-\frac{1}{12!}+...+\frac{1}{99!}-\frac{1}{100!}\)
\(=\frac{1}{9!}-\frac{1}{100!}< \frac{1}{9!}\left(đpcm\right)\)
Ta có
\(\frac{1}{n^{2}} + \frac{1}{n^{3}} = \frac{n + 1}{n^{3}} < \frac{2}{n^{2}} \leq \frac{1}{2^{\textrm{ } n - 1}} \left(\right. n \geq 2 \left.\right)\)Suy ra
\(A < \sum_{n = 2}^{1012} \frac{1}{2^{n - 1}} < \sum_{k = 1}^{\infty} \frac{1}{2^{k}} = 1\)Kết luận: \(A < 1\).
Sửa đề: \(A=\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{1013^2}\)
Ta có: \(\frac{1}{2^2}<\frac{1}{1\cdot2}=1-\frac12\)
\(\frac{1}{3^2}<\frac{1}{2\cdot3}=\frac12-\frac13\)
...
\(\frac{1}{1013^2}<\frac{1}{1012\cdot1013}=\frac{1}{1012}-\frac{1}{1013}\)
Do đó: \(\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{1013^2}<1-\frac12+\frac12-\frac13+\cdots+\frac{1}{1012}-\frac{1}{1013}\)
=>\(A<1-\frac{1}{1013}\)
=>A<1