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⇒ 1/k² < 1/[(k-1).(k+1)] = [1/(k-1) - 1/(k+1)]/2 (*)
Áp dụng (*), ta có:
1/2² + 1/3² + 1/4² + ... + 1/n²
< 1/2² + 1/(2.4) + 1/(3.5) + ... + 1/[(n-1).(n+1)]
= 1/2² + [1/2 - 1/4 + 1/3 - 1/5 + ... + 1/(n-1) - 1/(n+1)]/2
= 1/2² + [1/2 + 1/3 - 1/n - 1/(n+1)]/2
= 2/3 - [1/n + 1/(n+1)]/2 <2/3<1
Vì:\(\dfrac{1}{2^{^2}}>\dfrac{1}{1.2};\dfrac{1}{3^{^2}}>\dfrac{1}{2.3};\dfrac{1}{4^{^2}}>\dfrac{1}{3.4};...;\dfrac{1}{n^{^2}}>\dfrac{1}{\left(n-1\right)n}\)
\(\Rightarrow A>\dfrac{1 }{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{\left(n-1\right)n}\)
\(\Rightarrow A>\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{n-1}-\dfrac{1}{n}\)
\(\Rightarrow A>\dfrac{1}{1}-\dfrac{1}{n}\)
....
Tự lập luận làm tiếp
Ta có:
\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)
\(\dfrac{1}{4^2}< \dfrac{1}{3.4}\)
...
\(\dfrac{1}{n^2}< \dfrac{1}{n\left(n-1\right)}\)
\(\Rightarrow P< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{n\left(n-1\right)}\)
\(\Rightarrow P< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{n-1}-\dfrac{1}{n}\)
\(\Rightarrow P< 1-\dfrac{1}{n}< 1\)
\(\Rightarrow P< 1\)
Bài 2 :
\(S=\dfrac{1}{4}+\dfrac{2}{4^2}+\dfrac{3}{4^3}+............+\dfrac{2017}{4^{2017}}\)
\(\Leftrightarrow4S=1+\dfrac{2}{4}+\dfrac{3}{4^2}+...........+\dfrac{2017}{4^{2016}}\)
\(\Leftrightarrow4S-S=\left(1+\dfrac{2}{4}+\dfrac{3}{4^2}+..........+\dfrac{2017}{4^{2016}}\right)-\left(\dfrac{1}{4}+\dfrac{2}{4^2}+..........+\dfrac{2017}{4^{2017}}\right)\)
\(\Leftrightarrow3S=1+\dfrac{1}{4}+\dfrac{1}{4^2}+.........+\dfrac{1}{4^{2016}}-\dfrac{2017}{4^{2016}}\)
Đặt :
\(A=1+\dfrac{1}{4}+\dfrac{1}{4^2}+..........+\dfrac{1}{4^{2016}}\)
\(\Leftrightarrow4A=4+1+\dfrac{1}{4}+\dfrac{1}{4^2}+..........+\dfrac{1}{4^{2015}}\)
\(\Leftrightarrow4A-A=\left(4+1+\dfrac{1}{4}+.......+\dfrac{1}{4^{2015}}\right)-\left(1+\dfrac{1}{4}+.......+\dfrac{1}{4^{2016}}\right)\)
\(\Leftrightarrow3A=4-\dfrac{1}{4^{2016}}\)
\(\Leftrightarrow D=\dfrac{4}{3}-\dfrac{1}{2^{2016}.3}\)
\(\Leftrightarrow3S=\dfrac{4}{3}-\dfrac{1}{2^{2016}.3}-\dfrac{2017}{4^{2016}}\)
\(\Leftrightarrow3S< \dfrac{4}{3}\)
\(\Leftrightarrow S< \dfrac{4}{9}\)
\(\Leftrightarrow S< \dfrac{1}{2}\rightarrowđpcm\)
\(A=\dfrac{1}{4}+\dfrac{2}{4^2}+\dfrac{3}{4^3}+...+\dfrac{2017}{4^{2017}}\) ( A cho đẹp :v)
\(4A=4\left(\dfrac{1}{4}+\dfrac{2}{4^2}+\dfrac{3}{4^3}+...+\dfrac{2017}{4^{2017}}\right)\)
\(4A=1+\dfrac{2}{4}+\dfrac{3}{4^2}+...+\dfrac{2017}{4^{2016}}\)
\(4A-A=\left(1+\dfrac{2}{4}+\dfrac{3}{4^2}+...+\dfrac{2017}{4^{2016}}\right)-\left(\dfrac{1}{4}+\dfrac{2}{4^2}+\dfrac{3}{4^3}+...+\dfrac{2017}{4^{2017}}\right)\)\(3A=1+\dfrac{1}{4}+\dfrac{1}{4^2}+\dfrac{1}{4^3}+...+\dfrac{1}{4^{2016}}-\dfrac{2017}{4^{2017}}\)
Đặt:
\(M=1+\dfrac{1}{4}+\dfrac{1}{4^2}+\dfrac{1}{4^3}+...+\dfrac{1}{4^{2016}}\)
\(4M=4\left(1+\dfrac{1}{4}+\dfrac{1}{4^2}+\dfrac{1}{4^3}+...+\dfrac{1}{4^{2016}}\right)\)
\(4M=4+1+\dfrac{1}{4}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{2015}}\)
\(4M-M=\left(4+1+\dfrac{1}{4}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{2015}}\right)-\left(1+\dfrac{1}{4}+\dfrac{1}{4^2}+\dfrac{1}{4^3}+...+\dfrac{1}{4^{2016}}\right)\)\(3M=4-\dfrac{1}{4^{2016}}\)
\(M=\dfrac{4}{3}-\dfrac{1}{4^{2016}}\)
Thay M vào A ta có:
\(A=\dfrac{4}{9}-\dfrac{1}{4^{2016}.3}-\dfrac{2017}{4^{2017}}\)
\(\Rightarrow A< \dfrac{1}{2}\Rightarrowđpcm\)
\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{n\left(n-1\right)}\\ A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n-1}-\dfrac{1}{n}=1-\dfrac{1}{n}< 1\left(\dfrac{1}{n}>0\right)\)
Lời giải:
$n=1$ thì $S=0$ nguyên nhé bạn. Phải là $n>1$
\(S=1-\frac{1}{1^2}+1-\frac{1}{2^2}+1-\frac{1}{3^2}+...+1-\frac{1}{n^2}\)
\(=n-\underbrace{\left(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}\right)}_{M}\)
Để cm $S$ không nguyên ta cần chứng minh $M$ không nguyên. Thật vậy
\(M> 1+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{n(n+1)}=1+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{n}-\frac{1}{n+1}\)
\(M>1+\frac{1}{2}-\frac{1}{n+1}>1\) với mọi $n>1$
Mặt khác:
\(M< 1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{(n-1)n}=1+\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{n-1}-\frac{1}{n}\)
\(M< 1+1-\frac{1}{n}< 2\)
Vậy $1< M< 2$ nên $M$ không nguyên. Kéo theo $S$ không nguyên.
\(S=\dfrac{2}{2^1}+\dfrac{3}{2^2}+...+\dfrac{2017}{2^{2016}}\)
\(\Rightarrow2S=2+\dfrac{3}{2^1}+\dfrac{4}{2^2}+...+\dfrac{2017}{2^{2015}}\)
\(\Rightarrow2S-S=\left(2+\dfrac{3}{2^1}+\dfrac{4}{2^2}+...+\dfrac{2017}{2^{2015}}\right)-\left(\dfrac{2}{2^1}+\dfrac{3}{2^2}+...+\dfrac{2017}{2^{2016}}\right)\)
\(\Leftrightarrow S=2+\dfrac{1}{2^1}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2015}}-\dfrac{2017}{2^{2016}}\)
Tới đây thì đơn giản rồi nhé