Prove that 0.099<1/10^2+1/11^2+1/12^2+...+1/999^2+1/1000^2<0.111 without using any calculator or computer

Answer 1

I'll answer with whatever I know. This is just a theory of mine.
In geometric progression, the formula for sum to infinity is:

(a/(1-r)) where "a" is the first term and "r" is the common ratio between the terms.

In your case, there is no common ratio as the numbers are not decreasing proportionally. So I though if I can get the ratio for a few terms and get the average of the ratio to use in the formula.

This is what I got:

r for T1 and T2: 1/(11^2) / 1/(10^2)= (100/121)

r for T91 and T92: 1/(101^2) / 1/(100^2)= (10000/10201)
(have to do a bit of calculation here.)

get the average:
(100/121+10000/10201)/2 = it's around 0.9

so put into formula:

Sum to infinity= a/(1-r)
= (1./10^2)/(1-0.9)
= 0.1

so the Sum to infinity of the sequence 1/10^2, 1/11^2, 1/12^2,......
is equal to 0.1

so 0.099 is smaller than the sum to infinity which is equal to 0.1

I'm sure there are better ways but I think this could work.

Answer 2

if you work out the formula finally you have to prove that

sigma 1/(10+n)^2 [n=1 to infinity] > 0.089

at the time being I have no idea hove to prove this. I will come back in case I can find it.

Answer 3

If is microwave as in ovens, then there is not any want to even evaluate them because pcs would win. If the microwaves refers back to using magnetic pc transmission thingies, properly then through it self, microwaves are non usefull till used with the pcs for this reason. So, if it is the case, then pcs would win once again.

Answer 4

what ? I don't think you can