Đề là "+" thì may ra mk còn làm được, chứ "-" thì chưa nghĩ ra ._.
S = 1 + \(\frac{1}{2^2}\) + \(\frac{1}{3^2}\) + \(\frac{1}{4^2}\) + ... + \(\frac{1}{2020^2}\)
Ta có: 1 > \(\frac{1}{2020^2}\); \(\frac{1}{2^2}\) > \(\frac{1}{2020^2}\); \(\frac{1}{3^2}\) > \(\frac{1}{2020^2}\); ... ;\(\frac{1}{2019^2}\) > \(\frac{1}{2020^2}\) ; \(\frac{1}{2020^2}=\frac{1}{2020^2}\)
\(\Rightarrow\) 1 + \(\frac{1}{2^2}\) + \(\frac{1}{3^2}\) + \(\frac{1}{4^2}\) + ... + \(\frac{1}{2020^2}\) > \(\frac{1}{2020^2}\).2020
hay 1 + \(\frac{1}{2^2}\) + \(\frac{1}{3^2}\) + \(\frac{1}{4^2}\) + ... + \(\frac{1}{2020^2}\) > \(\frac{1}{2020}\) (đpcm)
Chúc bn học tốt!
