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phải là M<2/3 mới giải đc
\(M=\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+...+59}\)
\(=\frac{1}{\left(3+1\right).3:2}+\frac{1}{\left(4+1\right).4:2}+...+\frac{1}{\left(59+1\right).59:2}\)
\(=\frac{1}{6}+\frac{1}{10}+...+\frac{1}{1770}\)
\(=\frac{2}{12}+\frac{2}{20}+...+\frac{2}{3540}\)
\(=2\left(\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{595.60}\right)\)
\(=2\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{59}-\frac{1}{60}\right)\)
\(=2\left(\frac{1}{3}-\frac{1}{60}\right)\)
\(=\frac{2}{3}-\frac{2}{60}< \frac{2}{3}\)
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P=\(\left(1-\dfrac{1}{111}\right)\left(1-\dfrac{2}{111}\right)\times...\times\left(1-\dfrac{111}{111}\right)\times...\times\left(1-\dfrac{2009}{111}\right)\)
P=\(\left(1-\dfrac{1}{111}\right)\left(1-\dfrac{2}{111}\right)\times...\times0\times...\times\left(1-\dfrac{2009}{111}\right)\)
P=0
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\(M=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{10^2}\)
\(>1+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.11}\)
\(=1+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}\)
\(=1+\frac{1}{2}-\frac{1}{11}\)
\(>1+\frac{1}{2}-\frac{1}{6}=\frac{4}{3}\)