Find-the-sum-of-nth-term-k-1-n-1-k-2-

Question Number 78162 by TawaTawa last updated on 14/Jan/20

Find the sum of nth term

\underset{k = 1}{\sum^{n}} \frac{1}{k^{2}}

Commented by msup trace by abdo last updated on 15/Jan/20

i f y o u h a v e t h e a n s w e r p o s t i t \dots

Commented by TawaTawa last updated on 15/Jan/20

I {don}^{'} t know the answer or how to solve .

Commented by mathmax by abdo last updated on 15/Jan/20

t h i s s u m c a n b e r e d u c e d b y t h i s m a n n e r b y i a m w a i t i n g f o r

a m i r a c u l o u s m e t h o d . .! l e t ξ_{n} (2) = \sum_{k = 1}^{n} \frac{1}{k^{2}} \Rightarrow

ξ_{n} (2) = \sum_{p = 1}^{[\frac{n}{2}]} \frac{1}{{(2 p)}^{2}} + \sum_{p = 0}^{[\frac{n - 1}{2}]} \frac{1}{{(2 p + 1)}^{2}}

= \frac{1}{4} ξ_{[\frac{n}{2}]} (2) + \sum_{p = 0}^{[\frac{n - 1}{2}]} \frac{1}{{(2 p + 1)}^{2}} a n d

ξ_{[\frac{n}{2}]} (2) = \sum_{k = 1}^{[\frac{n}{2}]} \frac{1}{k^{2}} = \sum_{p = 1}^{[\frac{1}{2} [\frac{n}{2}]]} \frac{1}{{(2 p)}^{2}} + \sum_{p = 0}^{[\frac{[\frac{n}{2}] - 1}{2}]} \frac{1}{{(2 p + 1)}^{2}}

= ξ_{[\frac{[\frac{n}{2}]}{2}]} (2) + \sum_{p = 0}^{[\frac{[\frac{n}{2}] - 1}{2}]} \frac{1}{{(2 p + 1)}^{2}} a n d w e r e m p l a c e i n ξ_{n} (2)

a l s o w e c a n u s e t h e d e c o m p o s i t i o n

ξ_{n} (2) = \sum_{p = 1}^{[\frac{n}{3}]} \frac{1}{{(3 p)}^{2}} + \sum_{p = 0}^{[\frac{n - 1}{3}]} \frac{1}{{(3 p + 1)}^{2}} + \sum_{p = 0}^{[\frac{n - 2}{3}]} \frac{1}{{(3 p + 2)}^{2}}

= \dots .

Commented by TawaTawa last updated on 15/Jan/20

God bless you sir