Proving question, pattern and inequalities

Question

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

As you can see, I'm stuck please help D;

Answer 1

$$\frac1{1^2}+\frac1{2^2}+\frac1{3^2}+...+\frac1{1000^2}<$$ $$<\frac1{1^2}+\frac1{1\cdot2}+\frac1{2\cdot3}+...+\frac1{999\cdot1000}=$$ $$=1+\frac11-\frac12+\frac12-\frac13+...+\frac1{999}-\frac1{1000}=1+1-\frac1{1000}<2$$

Answer 2

By Riemann Sums:

$$\frac 1{1^2}+\frac 1{2^2}+\cdots +\frac 1{1000^2}=1+\left(\frac 1{2^2}+\cdots +\frac 1{1000^2}\right)≤1+\int_1^{1000}\frac {dx}{x^2}=1+\frac {999}{1000}<2$$