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Question Number 146735 by qaz last updated on 15/Jul/21

\underset{n = 1}{\sum^{\infty}} \frac{1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n}}{(n + 1) (n + 2)} = ?

Answered by Olaf_Thorendsen last updated on 15/Jul/21

S_{N} = \underset{n = 1}{\sum^{N}} \frac{H_{n}}{n + 1} - \underset{n = 1}{\sum^{N}} \frac{H_{n}}{n + 2}

S_{N} = \underset{n = 1}{\sum^{N}} \frac{H_{n} + \frac{1}{n + 1}}{n + 1} - \underset{n = 1}{\sum^{N}} \frac{1}{{(n + 1)}^{2}}

- \underset{n = 1}{\sum^{N}} \frac{H_{n} + \frac{1}{n + 1} + \frac{1}{n + 2}}{n + 2} + \underset{n = 1}{\sum^{N}} \frac{1}{(n + 1) (n + 2)}

+ \underset{n = 1}{\sum^{N}} \frac{1}{{(n + 2)}^{2}}

S_{N} = \underset{n = 2}{\sum^{N + 1}} \frac{H_{n}}{n} - \underset{n = 2}{\sum^{N + 1}} \frac{1}{n^{2}} - \underset{n = 3}{\sum^{N + 2}} \frac{H_{n}}{n}

+ \underset{n = 2}{\sum^{N + 1}} \frac{1}{n} - \underset{n = 3}{\sum^{N + 2}} \frac{1}{n} + \underset{n = 3}{\sum^{N + 2}} \frac{1}{n^{2}}

S_{N} = \frac{H_{2}}{2} - \frac{H_{n + 2}}{n + 2} + \frac{1}{{(n + 2)}^{2}} - \frac{1}{2^{2}} + \frac{1}{2} - \frac{1}{n + 2}

H_{n} \sim \ln n + γ \Rightarrow \frac{H_{n + 2}}{n + 2} \to 0

Finally S_{\infty} = \frac{1 + \frac{1}{2}}{2} - \frac{1}{4} + \frac{1}{2} = 1

Answered by mnjuly1970 last updated on 15/Jul/21

s o l u t i o n . .

w e k n o w t h a t :

\sum_{n ⩾ 1} \frac{H_{n} x^{n + 1}}{n + 1} = \frac{1}{2} {l o g}^{2} (1 - x)

\sum_{n ⩾ 1} \frac{H_{n}}{(n + 1) (n + 2)} = \frac{1}{2} \int_{0}^{1} {l o g}^{2} (x) d x

= \frac{1}{2} {{[{x l o g}^{2} (x)]}_{0}^{1} - 2 \int_{0}^{1} l o g (x) d x}

= - \int_{0}^{1} l o g (x) d x = 1 \dots . ✓

Commented by qaz last updated on 15/Jul/21

\underset{n = 1}{\sum^{\infty}} \frac{H_{n}}{(n + 1) (n + 2)} = \underset{n = 1}{\sum^{\infty}} H_{n} \int_{0}^{1} (x^{n} - x^{n + 1}) dx = \int_{0}^{1} (- \frac{\ln (1 - x)}{1 - x} + \frac{xln (1 - x)}{1 - x}) dx

= - \int_{0}^{1} \ln (1 - x) dx = 1

- - - - - - - - - - - - - - -

\underset{n = 1}{\sum^{\infty}} H_{n} x^{n} = \underset{n = 1}{\sum^{\infty}} \underset{k = 1}{\sum^{n}} \frac{x^{n}}{k} = \underset{k = 1}{\sum^{\infty}} \underset{n = k}{\sum^{\infty}} \frac{x^{n}}{k} = \underset{k = 1}{\sum^{\infty}} \underset{n = 0}{\sum^{\infty}} \frac{x^{n + k}}{k} = (\underset{k = 1}{\sum^{\infty}} \frac{x^{k}}{k}) (\underset{n = 0}{\sum^{\infty}} x^{n}) = - \frac{\ln (1 - x)}{1 - x}

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