n-1-1-n-2-

Question Number 32755 by 7991 last updated on 01/Apr/18

\underset{n = 1}{\sum^{\infty}} \frac{1}{n^{2}} = \dots . ? ? ?

Commented by Rio Mike last updated on 01/Apr/18

solution

n = 1 \Rightarrow \frac{1}{1^{2}}

n = 2 \Rightarrow \frac{1}{2^{2}}

n = 3 \Rightarrow \frac{1}{2^{3}}

the GP is 1, \frac{1}{4}, \frac{1}{8}, \dots

S_{\infty} = \frac{a}{1 - r}

\Rightarrow \frac{1}{1 - \frac{1}{4}}

= 1 \times \frac{4}{3}

= \frac{4}{3}

Commented by abdo imad last updated on 01/Apr/18

\sum_{n = 1}^{\infty} \frac{1}{n^{2}} = ξ (2) = \frac{π^{2}}{6} w i t h ξ (x) = \sum_{n = 1}^{\infty} \frac{1}{n^{x}} w i t h x > 1

y o u w i l l f i n d t h e p r o o f i n t h i s p l a t f o r m .

Commented by abdo imad last updated on 01/Apr/18

y o u r a n s w e r i s n o t c o r r e c t R i o s i r \dots .

Commented by JDamian last updated on 05/Apr/18

{I s n}^{'} t t h i s k n o w n a s {B a s e l}^{'} s p r o b l e m ?

Commented by Rasheed.Sindhi last updated on 05/Apr/18

1, \frac{1}{4}, \frac{1}{8}, \dots is not GP

∵ \frac{1}{4} \div 1 \neq \frac{1}{8} \div \frac{1}{4}

\frac{1}{4} \neq \frac{1}{2}