nice-calculus-evaluate-n-1-3n-1-3-

Question Number 138223 by mnjuly1970 last updated on 11/Apr/21

\dots \dots . n i c e \dots \dots \dots c a l c u l u s \dots

e v a l u a t e :::

Θ = \underset{n = - \infty}{\sum^{\infty}} \frac{1}{{(3 n + 1)}^{3}} = ?

\dots \dots \dots \dots \dots \dots \dots \dots .

Answered by Dwaipayan Shikari last updated on 11/Apr/21

\sum_{n \in Z} \frac{1}{{(3 n + 1)}^{3}} = \underset{n = - \infty}{\sum^{\infty}} \frac{1}{{(3 n + 1)}^{3}} - 1

\underset{n = - \infty}{\sum^{\infty}} \frac{1}{(x + π n)} = c o t x \Rightarrow \underset{n = - \infty}{\sum^{\infty}} \frac{1}{{(x + π n)}^{2}} = \frac{1}{{s i n}^{2} x} = 1 + {c o t}^{2} x

\underset{n = - \infty}{\sum^{\infty}} \frac{2}{{(x + π n)}^{3}} = 2 c o t x {c o s e c}^{2} x

\underset{n = - \infty}{\sum^{\infty}} \frac{1}{{(\frac{π}{3} + π n)}^{3}} = c o t (\frac{π}{3}) {c o s e c}^{2} (\frac{π}{3})

\Rightarrow \underset{n = - \infty}{\sum^{\infty}} \frac{1}{{(3 n + 1)}^{3}} = \frac{π^{3}}{27} (\frac{1}{\sqrt{3}} . \frac{4}{3}) \Rightarrow \underset{n = - \infty}{\sum^{\infty}} \frac{1}{{(3 n + 1)}^{3}} = \frac{4 π^{3}}{81 \sqrt{3}}

\sum_{n \in Z} \frac{1}{{(3 n + 1)}^{3}} = \frac{4 π^{3}}{81 \sqrt{3}} - 1

Answered by mnjuly1970 last updated on 11/Apr/21

t h a n k y o u s o m u c h \dots

Commented by Dwaipayan Shikari last updated on 11/Apr/21

S o r r y i h a d a t y p o . k i n d l y c h e c k

Commented by mnjuly1970 last updated on 11/Apr/21

y o u a r e w e l c o m

y o u r a n s w e r i s c o r r e c t . m e r c e y

\underset{n = - \infty}{\sum^{\infty}} \frac{1}{{(3 n + 1)}^{2}} = \frac{4 π^{3} \sqrt{3}}{243} \dots ✓