n-1-1-1-2-1-n-1-2-n-ln4-

Question Number 104604 by ~blr237~ last updated on 22/Jul/20

\underset{n = 1}{\sum^{\infty}} (1 + \frac{1}{2} + \dots + \frac{1}{n}) \frac{1}{2^{n}} = l n 4

Answered by OlafThorendsen last updated on 22/Jul/20

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

S = \underset{n = 1}{\sum^{\infty}} \frac{1}{2^{n}} \underset{k = 1}{\sum^{n}} \frac{1}{k}

S = \underset{n = 1}{\sum^{\infty}} \frac{1}{n} \underset{k = n}{\sum^{\infty}} \frac{1}{2^{k}}

S = \lim_{N \to \infty} \underset{n = 1}{\sum^{\infty}} \frac{1}{n} . \frac{1}{2^{n}} . \frac{1 - {(\frac{1}{2})}^{N}}{1 - \frac{1}{2}}

S = 2 \underset{n = 1}{\sum^{\infty}} \frac{1}{n} . \frac{1}{2^{n}}

S = 2 \underset{n = 1}{\sum^{\infty}} \frac{2^{- n}}{n}

\frac{1}{1 - x} = \underset{n = 0}{\sum^{\infty}} x^{k}, ∣ x ∣< 1

- \ln ∣ 1 - x ∣ = \underset{n = 1}{\sum^{\infty}} \frac{x^{k}}{k}

S = 2 \underset{n = 1}{\sum^{\infty}} \frac{2^{- n}}{n} = - 2 \ln ∣ 1 - \frac{1}{2} ∣

S = 2 \underset{n = 1}{\sum^{\infty}} \frac{2^{- n}}{n} = - 2 \ln \frac{1}{2} = \ln 4

Commented by ~blr237~ last updated on 23/Jul/20

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