calculate-U-n-k-0-n-1-3k-1-interms-of-H-n-k-1-n-1-k-

Question Number 83214 by mathmax by abdo last updated on 28/Feb/20

c a l c u l a t e U_{n} = \sum_{k = 0}^{n} \frac{1}{3 k + 1} i n t e r m s o f H_{n} = \sum_{k = 1}^{n} \frac{1}{k}

Answered by ~blr237~ last updated on 29/Feb/20

l e t n ⩾ 1

H_{3 n} = \underset{k = 1}{\sum^{3 n}} \frac{1}{k} = \underset{k = 1}{\sum^{n}} \frac{1}{3 k} + \underset{k = 0}{\sum^{n - 1}} \frac{1}{3 k + 1} + \underset{k = 0}{\sum^{n - 1}} \frac{1}{3 k + 2}

= \frac{1}{3} H_{n} + U_{n - 1} + V_{n - 1} (1) w h e r e V_{n} = \underset{k = 0}{\sum^{n}} \frac{1}{3 k + 2}

H_{3 n + 1} = \underset{k = 1}{\sum^{n}} \frac{1}{3 k} + \underset{k = 0}{\sum^{n}} \frac{1}{3 k + 1} + \underset{k = 0}{\sum^{n - 1}} \frac{1}{3 k + 2}

= \frac{1}{3} H_{n} + U_{n} + V_{n - 1} (2)

(2) - (1) \Rightarrow U_{n} - U_{n - 1} = H_{3 n + 1} - H_{3 n}

s o \underset{k = 1}{\sum^{n}} (U_{k} - U_{k - 1}) = \underset{k = 1}{\sum^{n}} (H_{3 k + 1} - H_{3 k})

T h e n U_{n} = U_{0} + \underset{k = 1}{\sum^{n}} (H_{3 k + 1} - H_{3 k})

Facebook Tweet Pin