If-a-1-2-2-1-3-2-1-100-2-b-0-99-Prove-that-a-lt-b-

Question Number 204545 by hardmath last updated on 21/Feb/24

If

a = \frac{1}{2^{2}} + \frac{1}{3^{2}} + \dots + \frac{1}{100^{2}}

b = 0, 99

Prove that :

a < b

Answered by mr W last updated on 21/Feb/24

a = \frac{1}{2^{2}} + \frac{1}{3^{2}} + \dots + \frac{1}{100^{2}}

= (1 + \frac{1}{2^{2}} + \frac{1}{3^{2}} + \dots + \frac{1}{100^{2}} + \frac{1}{101^{2}} + \dots) - 1 - (\frac{1}{101^{2}} + \dots)

< (1 + \frac{1}{2^{2}} + \frac{1}{3^{2}} + \dots + \frac{1}{100^{2}} + \frac{1}{101^{2}} + \dots) - 1

= \frac{π^{2}}{6} - 1 < 0.99

Answered by witcher3 last updated on 21/Feb/24

k^{2} ⩾ k (k - 1); \forall k ⩾ 2

\Rightarrow \frac{1}{k^{2}} ⩽ \frac{1}{k (k - 1)} \Rightarrow a = \underset{k = 2}{\sum^{100}} \frac{1}{k^{2}} ⩽ \underset{k = 2}{\sum^{100}} \frac{1}{k (k - 1)} = \underset{k = 2}{\sum^{100}} (\frac{1}{k - 1} - \frac{1}{k}) = 1 - \frac{1}{100} = 0.99

a < b

Commented by mr W last updated on 21/Feb/24

g r e a t a p p r o a c h!

b u t w i t h t h i s m e t h o d w e c a n n o t p r o v e

\frac{1}{2^{2}} + \frac{1}{3^{2}} + \dots + \frac{1}{100^{2}} < \frac{3}{4}

Commented by witcher3 last updated on 22/Feb/24

exactly same aproch will not Give this

but i think That its to prove < \frac{3}{4}

using interpelation of lagrange

approching f (x) = \frac{1}{x^{2}}; withe polynome

Commented by mr W last updated on 22/Feb/24

t h a n k s a l o t!

Commented by hardmath last updated on 23/Feb/24

thank you dear professors