Calculus-n-1-1-n-k-1-n-k-2-

Question Number 143603 by mnjuly1970 last updated on 16/Jun/21

\dots . . C a l c u l u s \dots . .

Ω := \underset{n = 1}{\sum^{\infty}} \frac{1}{n^{k} (1 + n)} (k ⩾ 2) \dots \dots

Answered by Dwaipayan Shikari last updated on 16/Jun/21

Ω := \underset{n = 1}{\sum^{\infty}} \frac{1}{n^{k} (n + 1)} = \int_{0}^{1} {L i}_{k} (x) d x

= {[{x L i}_{k} (x)]}_{0}^{1} - \int_{0}^{1} \underset{n = 1}{\sum^{\infty}} n \frac{x^{n - 1} . x}{n^{k}} d x

= {L i}_{k} (1) - \int_{0}^{1} \underset{n = 1}{\sum^{\infty}} \frac{x^{n}}{n^{k - 1}} d x = {L i}_{k} (1) - \int_{0}^{1} {L i}_{k - 1} (1) d x

= {L i}_{k} (1) - {L i}_{k - 1} (1) + {L i}_{k - 2} (1) - \dots {L i}_{2} (1)

= ζ (k) - ζ (k - 1) + ζ (k - 2) - ζ (k - 3) + \dots + \frac{π^{2}}{6} i f k o d d

o r ζ (k) - ζ (k - 1) + \dots - \frac{π^{2}}{6} i f k e v e n

Commented by mnjuly1970 last updated on 16/Jun/21

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

Answered by mindispower last updated on 16/Jun/21

f (k) = \sum_{n ⩾ 1} \frac{1}{n^{k} (n + 1)}, f (1) = 1

f (k) + f (k - 1) = Σ \frac{1}{n^{k}} = ζ (k), k ⩾ 1

{(- 1)}^{m} f (m) + {(- 1)}^{m} f (m - 1) = {(- 1)}^{m} ζ (m)

\underset{m = 2}{\sum^{k}} {(- 1)}^{m} f (m) + {(- 1)}^{m} f (m - 1) = \underset{m = 2}{\sum^{k}} {(- 1)}^{m} ζ (m)

{(- 1)}^{k} f (k) + f (1) = \underset{m = 2}{\sum^{k}} {(- 1)}^{m} ζ (m)

f (k) = {(- 1)}^{k} \underset{m = 2}{\sum^{k}} {(- 1)}^{m} ζ (m) + {(- 1)}^{k - 1}, k ⩾ 2

f (1) = 1

Commented by mnjuly1970 last updated on 16/Jun/21

v e r y n i c e \dots .

Commented by mindispower last updated on 16/Jun/21

p l e a s u r