\(\frac{1}{2^2}\)+\(\frac{1}{3^2}\)+...+\(\frac{1}{50^2}\)<1
ta có \(\frac{1}{2^2}\)<\(\frac{1}{1.2}\)
\(\frac{1}{3^2}\)<\(\frac{1}{2.3}\)
..........................
\(\frac{1}{50^2}\)<\(\frac{1}{49.50}\)
ta được \(\frac{1}{1.2}\)+\(\frac{1}{2.3}\)+...+\(\frac{1}{49.50}\)
=>1-\(\frac{1}{2}\)+\(\frac{1}{2}\)-...-\(\frac{1}{49}\)+\(\frac{1}{49}\)-\(\frac{1}{50}\)
=>1-\(\frac{1}{50}\)<1 nên\(\frac{1}{2^2}\)+\(\frac{1}{3^2}\)+...+\(\frac{1}{50^2}\)<1
vậy ...........................