Yahir Tucker

2022-06-28

Proving question, pattern and inequalities
Hi my math tutor gave me this problem to do over the week: prove 1/1^2+1/2^2+1/3^2...1/1000^2 <2
I've managed to almost complete the question, but I think I'm missing something or made a mistake.
My attempt at the question: 1/1000^2 < 1/999*1000 1/2^2+1/3^2+...1/1000^2<1/1*2+1/2*3+...1/999*1000 1/1*2+1/2*3+...1/999*1000= (1-1/2)+(1/2-1/3)+(1/3-1/4)...+1/999-1/1000 (we can cancel out each fraction except for 1-1/1000)
Therefore, 1/1*2+1/2*3+...1/999*1000=1-1/1000 1/2^2+1/3^2+...1/1000^2<1-1000 1/1^2+1/2^2+1/3^2...1/1000^2<2-1000

$\frac{1}{{1}^{2}}+\frac{1}{{2}^{2}}+\frac{1}{{3}^{2}}+...+\frac{1}{{1000}^{2}}<$
$<\frac{1}{{1}^{2}}+\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+...+\frac{1}{999\cdot 1000}=$
$=1+\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{999}-\frac{1}{1000}=1+1-\frac{1}{1000}<2$

Lydia Carey

By Riemann Sums:
$\frac{1}{{1}^{2}}+\frac{1}{{2}^{2}}+\cdots +\frac{1}{{1000}^{2}}=1+\left(\frac{1}{{2}^{2}}+\cdots +\frac{1}{{1000}^{2}}\right)\le 1+{\int }_{1}^{1000}\frac{dx}{{x}^{2}}=1+\frac{999}{1000}<2$

Do you have a similar question?