find-lim-n-1-i-lt-j-n-1-ij-2-

Question Number 72023 by mathmax by abdo last updated on 23/Oct/19

f i n d {l i m}_{n \to + \infty} \sum_{1 ⩽ i < j ⩽ n} \frac{1}{{(i j)}^{2}}

Commented by mathmax by abdo last updated on 24/Oct/19

w e h a v e {(\sum_{i = 1}^{n} x_{i})}^{2} = \sum_{i = 1}^{n} x_{i}^{2} + 2 \sum_{1 ⩽ i < j ⩽ n} x_{i} x_{j}

l e t x_{i} = \frac{1}{i^{2}} \Rightarrow {(\sum_{i = 1}^{n} \frac{1}{i^{2}})}^{2} = \sum_{i = 1}^{n} \frac{1}{i^{4}} + 2 \sum_{1 ⩽ i < j ⩽ n} \frac{1}{i^{2}} \times \frac{1}{j^{2}}

\Rightarrow {l i m}_{n \to + \infty} {(\sum_{i = 1}^{n} \frac{1}{i^{2}})}^{2} = {l i m}_{n \to + \infty} \sum_{i = 1}^{n} \frac{1}{i^{4}} + 2 {l i m}_{n \to + \infty} \sum_{1 ⩽ i < j ⩽ n} \frac{1}{{(i j)}^{2}} \Rightarrow

2 {l i m}_{n \to + \infty} \sum_{1 ⩽ i < j ⩽ n} \frac{1}{{(i j)}^{2}} = {(\sum_{i = 1}^{\infty} \frac{1}{i^{2}})}^{2} - \sum_{n = 1}^{\infty} \frac{1}{i^{4}}

= {(\frac{π^{2}}{6})}^{2} - ξ (4) = \frac{π^{4}}{36} - ξ (4) \Rightarrow

{l i m}_{n \to + \infty} \sum_{1 ⩽ i < j ⩽ n} \frac{1}{{(i j)}^{2}} = \frac{π^{4}}{72} - \frac{ξ (4)}{2}

Answered by mind is power last updated on 23/Oct/19

\sum_{1 ⩽ i < j ⩽ n} = \underset{j = 2}{\sum^{n}} \frac{1}{j^{2}} \underset{i = 1}{\sum^{j - 1}} \frac{1}{i^{2}}

\sum_{1 ⩽ j < i ⩽ n} \frac{1}{{(ij)}^{2}} = \sum_{1 ⩽ i < j ⩽ n} \frac{1}{{(ij)}^{2}}

\underset{j, i = 1}{\sum^{n}} \frac{1}{{(ij)}^{2}} = \underset{i = 1}{\sum^{n}} \frac{1}{i^{2}} . \underset{j = 1}{\sum^{n}} \frac{1}{j^{2}}

\sum_{1 ⩽ j < i ⩽ n} \frac{1}{{(ij)}^{2}} + \sum_{1 ⩽ i < j ⩽ n} \frac{1}{{(ij)}^{2}} + \underset{i = j = 1}{\sum^{n}} \frac{1}{{(ij)}^{2}} + \underset{i = 1}{\sum^{n}} \frac{1}{i^{2}} \underset{j = 1}{\sum^{n}} \frac{1}{j^{2}}

2 \sum_{1 ⩽ i < j ⩽ n} \frac{1}{{(ij)}^{2}} = \underset{i = 1}{\sum^{n}} \frac{1}{i^{2}} . \underset{j . 1}{\sum^{n}} \frac{1}{j^{2}} - \underset{1}{\sum^{n}} \frac{1}{j^{4}}

2 \sum_{1 ⩽ i < j ⩽ n} \frac{1}{{(ij)}^{2}} = {(ζ (2))}^{2} - ζ (4) \Rightarrow

\sum_{1 ⩽ i < j ⩽ n} \frac{1}{{(ij)}^{2}} = \frac{ζ {(2)}^{2} - ζ (4)}{2}

Commented by gunawan last updated on 23/Oct/19

you are Amazing

Commented by mind is power last updated on 23/Oct/19

thanx sir