n-k-1-n-1-k-1-n-k-H-k-Find-the-value-of-n-1-1-n-1-n-2-

Question Number 156386 by mnjuly1970 last updated on 10/Oct/21

ϕ (n) = \underset{k = 1}{\sum^{n}} {(- 1)}^{k - 1} (\begin{matrix} n \\ k \end{matrix}) H_{k}

Find the value of :

\underset{n = 1}{\sum^{\infty}} {(- 1)}^{n - 1} ϕ (n^{2}) = ?

Answered by mindispower last updated on 10/Oct/21

1 = \emptyset (1)

2 - (1 + \frac{1}{2}) = \frac{1}{2} = \emptyset (2)

ϕ (3) = 3 - 3 (\frac{3}{2}) + (1 + \frac{1}{2} + \frac{1}{3}) = \frac{1}{3} = \emptyset (3)

ϕ (n) = \frac{1}{n} . . ?

H_{k} = \int_{0}^{1} \frac{1 - x^{k}}{1 - x} d x

\underset{k = 1}{\sum^{n}} \int_{0}^{1} {(- 1)}^{k - 1} (\begin{matrix} n \\ k \end{matrix}) \frac{1 - x^{k}}{1 - x} d x = \emptyset (n)

= \int_{0}^{1} \frac{1}{1 - x} + \frac{{(1 - x)}^{n} - 1}{1 - x} d x

= \int_{0}^{1} \frac{{(1 - x)}^{n}}{1 - x} d x = \frac{1}{n} = ϕ (n)

\sum_{n ⩾ 1} \frac{{(- 1)}^{n - 1}}{n^{2}} = - {L i}_{2} (- 1) = \frac{π^{2}}{12}

Commented by mnjuly1970 last updated on 10/Oct/21

v e r y n i c e . . e x c e l l e n t . . t h a n k y o u

s o m u c h s i r p o w e r

Commented by mindispower last updated on 10/Oct/21

w i t h e p l e a s u r s i r