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Question Number 194559 by MM42 last updated on 09/Jul/23

r e p e a t q u e s t i o n

S h i w t h a t :

\underset{i = 1}{\sum^{n}} (\frac{1}{2 i - 1} - \frac{1}{2 i}) = \underset{i = 1}{\sum^{n}} \frac{1}{n + i} ?

Answered by witcher3 last updated on 10/Jul/23

n = 1 \Rightarrow 1 - \frac{1}{2} = \frac{1}{2} = \frac{1}{1 + 1}

\forall n \in Z_{+} suppose True \underset{i = 1}{\sum^{n}} \frac{1}{2 i - 1} - \frac{1}{2 i} = Σ \frac{1}{n + i}

\underset{i = 1}{\sum^{n + 1}} \frac{1}{2 i - 1} - \frac{1}{2 i} \overset{?}{=} \underset{i = 1}{\sum^{n}} \frac{1}{n + 1 + i} = \underset{i = 1}{\sum^{n - 1}} \frac{1}{n + 1 + i} + \frac{1}{2 n + 1}

\Leftrightarrow \underset{i = 1}{\sum^{n}} (\frac{1}{2 i - 1} - \frac{1}{2 i}) + \frac{1}{2 n + 1} - \frac{1}{2 n + 2} =

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

= \underset{i = 1}{\sum^{n - 1}} \frac{1}{n + 1 + i} + \frac{1}{n + 1 + n} + \frac{1}{n + 1 + n + 1}

= \underset{i = 1}{\sum^{n + 1}} \frac{1}{(n + 1) + i} \dots . True

Commented by MM42 last updated on 10/Jul/23

e x c e l l e n t ★

Commented by witcher3 last updated on 11/Jul/23

thank You sir

Answered by MM42 last updated on 10/Jul/23

(1 - \frac{1}{2}) + (\frac{1}{3} - \frac{1}{4}) + \dots + (\frac{1}{2 n - 1} - \frac{1}{2 n})

= (1 + \frac{1}{3} + \dots + \frac{1}{2 n - 1}) - (\frac{1}{2} + \frac{1}{4} + \dots + \frac{1}{2 n})

+ 2 (\frac{1}{2} + \frac{1}{4} + \dots + \frac{1}{2 n}) - 2 (\frac{1}{2} + \frac{1}{4} + \dots + \frac{1}{2 n})

= (1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n} + \frac{1}{n + 1} + \dots + \frac{1}{2 n}) - (1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n})

= \frac{1}{n + 1} + \frac{1}{n + 2} + \dots + \frac{1}{2 n} ✓

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