Xét thừa số tổng quát: \(n!=1.2.3...n\)
Ta có:
\(L=\dfrac{1}{2!}+\dfrac{2}{3!}+\dfrac{3}{4!}+...+\dfrac{99}{100!}\)
\(L=\dfrac{2-1}{2!}+\dfrac{3-1}{3!}+\dfrac{4-1}{4!}+...+\dfrac{100-1}{100!}\)
\(L=\dfrac{2-1}{1.2}+\dfrac{3-1}{1.2.3}+\dfrac{4-1}{1.2.3.4}+...+\dfrac{100-1}{1.2.3...100}\)
\(L=1-\dfrac{1}{1.2}+\dfrac{1}{1.2}-\dfrac{1}{1.2.3}+\dfrac{1}{1.2.3}-\dfrac{1}{1.2.3.4}+...+\dfrac{1}{1.2.3....99}-\dfrac{1}{1.2.3...100}\)
\(L=1-\dfrac{1}{1.2.3....100}< 1\left(đpcm\right)\)
Xét thừa số tổng quát: n ! = 1.2.3 ... nn!=1.2.3 ...n
Ta có:
L = 1 2 !+ 2 3 !+ 3 4 !+ . . . + 99 100 !L=12!+23!+34!+...+99100!
L = 2 - 1 2 !+ 3 - 1 3 !+ 4 - 1 4 !+ . . . + 100 - 1 100 !L=2- -12!+3- -13!+4- -14!+...+100- -1100!
L = 2 - 1 1.2+ 3 - 1 1.2.3+ 4 - 1 1.2.3.4+ . . . + 100 - 1 1.2.3 ... 100L=2−11.2+3−11.2.3+4−11.2.3.4+...+100−11.2.3...100
L = 1 - 1 1.2+ 1 1.2- 1 1.2.3+ 1 1.2.3- 1 1.2.3.4+ . . . + 1 1.2.3 .... 99- 1 1.2.3 ... 100L=1−11.2+11.2−11.2.3+11.2.3−11.2.3.4+...+11.2.3....99−11.2.3 ... 100
L = 1 - 1 1.2.3 .... 100<1(đpcm)
