Question

Chứng minh \(\dfrac{1}{1+2+3}\)+\(\dfrac{1}{1+2+3+4}\)+......+\(\dfrac{1}{1+2+3+4+...+59}\)<\(\dfrac{2}{3}\)

 

Answer 1

\(\dfrac{1}{1+2+3+...+n}=\dfrac{1}{\dfrac{n\left(n+1\right)}{2}}=\dfrac{2}{n\left(n+1\right)}=\dfrac{2}{n}-\dfrac{2}{n+1}\)

Do đó:

\(\dfrac{1}{1+2+3}+\dfrac{1}{1+2+3+4}+...+\dfrac{1}{1+2+...+59}=\dfrac{2}{3}-\dfrac{2}{4}+\dfrac{2}{4}-\dfrac{2}{5}+...+\dfrac{2}{59}-\dfrac{2}{60}\)

\(=\dfrac{2}{3}-\dfrac{2}{60}< \dfrac{2}{3}\) (đpcm)