For-S-1-1-2-1-3-1-n-S-H-n-Harmonic-sequence-H-n-i-1-n-1-i-Can-you-solve-the-partial-sum-

Question Number 3929 by Filup last updated on 25/Dec/15

For :

S = 1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n}

S = H_{n} H a r m o n i c s e q u e n c e

H_{n} = \underset{i = 1}{\sum^{n}} \frac{1}{i}

C a n y o u s o l v e t h e p a r t i a l s u m ?

Commented by Yozzii last updated on 25/Dec/15

W o l f r a m A l p h a g i v e s

H_{n} = ψ_{0} (n + 1) + γ

w h e r e γ - E u l e r M a s c h e r o n i c o n s t a n t

ψ_{0} (z) i s t h e d i g a m m a f u n c t i o n d e f i n e d

b y t h e l o g a r i t h m i c d e r i v a t i v e o f t h e

g a m m a f u n c t i o n Γ (z) .

ψ_{0} (z) = \frac{d}{d z} {l n Γ (z)} = \frac{Γ^{^{'}} (z)}{Γ (z)}

T h e d i g a m m a f u n c t i o n s a t i s f i e s

ψ_{0} (z) = \int_{0}^{\infty} (\frac{e^{- t}}{t} - \frac{e^{- z t}}{1 - e^{- t}}) d t .

F o r z \equiv n, n \in Z^{+},

ψ_{0} (n) = - γ + \underset{i = 1}{\sum^{n - 1}} \frac{1}{i} \Rightarrow H_{n - 1} = ψ_{0} (n) + γ .

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