calculate-lim-n-1-i-lt-j-n-1-i-x-j-x-with-x-gt-1-for-that-consider-x-n-1-1-n-x-2-calculate-lim-n-1-i-lt-j-n-1-ij-2-

Question Number 42493 by maxmathsup by imad last updated on 26/Aug/18

c a l c u l a t e {l i m}_{n \to + \infty} \sum_{1 ⩽ i < j ⩽ n} \frac{1}{i^{x} j^{x}} w i t h x > 1 f o r t h a t c o n s i d e r

ξ (x) = \sum_{n = 1}^{\infty} \frac{1}{n^{x}}

2) c a l c u l a t e {l i m}_{n \to + \infty} \sum_{1 ⩽ i < j ⩽ n} \frac{1}{{(i j)}^{2}} .

Commented by maxmathsup by imad last updated on 29/Aug/18

l e t S_{n} (x) = \sum_{1 ⩽ i < j ⩽ n} \frac{1}{i^{x} j^{x}} a n d ξ_{n} (x) = \sum_{k = 1}^{n} \frac{1}{k^{x}} w e h a v e

{(\sum_{k = 1}^{n} \frac{1}{k^{x}})}^{2} = \sum_{k = 1}^{n} \frac{1}{k^{2 x}} + 2 \sum_{1 ⩽ i < j ⩽ n} \frac{1}{i^{x} j^{x}} l e t p a s s e t o l i m i t (n \to + \infty)

w e g e t {(ξ (x))}^{2} = ξ (2 x) + 2 {l i m}_{n \to + \infty} S_{n} (x) \Rightarrow

{l i m}_{n \to + \infty} S_{n} (x) = \frac{1}{2} {{(ξ (x))}^{2} - ξ (2 x)}

2) w e h a v e {l i m}_{n \to + \infty} \sum_{1 ⩽ i < j ⩽ n} \frac{1}{i^{2} j^{2}} = {l i m}_{n \to + \infty} S_{n} (2)

= \frac{1}{2} {{(ξ (2))}^{2} - ξ (4)} = \frac{1}{2} {{(\frac{π^{2}}{6})}^{2} - ξ (4)}

= \frac{1}{2} {\frac{π^{4}}{36} - ξ (4)} .