S_n =Σ_(n=1) ^n (1/2^k )tanh((1/2^k ))=?


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Question Number 144272 by SOMEDAVONG last updated on 24/Jun/21

$S_{n} = \underset{n = 1}{\sum^{n}} \frac{1}{2^{k}} \tanh (\frac{1}{2^{k}}) = ?$
	Answered by Olaf_Thorendsen last updated on 24/Jun/21

	$S_{n} = \underset{k = 1}{\sum^{n}} \frac{1}{2^{k}} \tanh (\frac{x}{2^{k}})$ $f_{n} (x) = \underset{k = 1}{\prod^{n}} \cosh (\frac{x}{2^{k}}) (1)$ $\sinh 2 u = 2 \cosh u . sinhh u$ $\cosh u = \frac{\sinh 2 u}{2 \sinh u}$ $f_{n} (x) = \underset{k = 1}{\prod^{n}} \frac{\sinh (\frac{x}{2^{k - 1}})}{2 \sinh (\frac{x}{2^{k}})}$ $f_{n} (x) = \frac{1}{2^{n}} . \frac{\sinh (x)}{\sinh (\frac{x}{2^{n}})} (2)$ $(1) : \ln f_{n} (x) = \ln \underset{k = 1}{\prod^{n}} \cosh (\frac{x}{2^{k}})$ $\ln f_{n} (x) = \underset{k = 1}{\sum^{n}} \ln (\cosh (\frac{x}{2^{k}}))$ $\frac{d}{d x} \ln f_{n} (x) = \underset{k = 1}{\sum^{n}} \frac{1}{2^{k}} \tanh (\frac{x}{2^{k}})$ $\frac{d}{d x} \ln f_{n} (1) = \underset{k = 1}{\sum^{n}} \frac{1}{2^{k}} \tanh (\frac{1}{2^{k}}) = S_{n} (3)$ $(2) : \ln f_{n} (x) = \ln (\sinh x) - \ln (\sinh (\frac{x}{2^{n}})) - n \ln 2$ $\frac{d}{d x} \ln f_{n} (x) = \coth x - \frac{1}{2^{n}} \coth (\frac{x}{2^{n}})$ $\frac{d}{d x} \ln f_{n} (1) = \coth (1) - \frac{1}{2^{n}} \coth (\frac{1}{2^{n}}) (4)$ $(3) and (4) :$ $S_{n} = \coth (1) - \frac{1}{2^{n}} \coth (\frac{1}{2^{n}})$ $S_{n} = \frac{e + e^{- 1}}{e - e^{- 1}} - \frac{1}{2^{n}} . \frac{e^{\frac{1}{2^{n}}} + e^{- \frac{1}{2^{n}}}}{e^{\frac{1}{2^{n}}} - e^{- \frac{1}{2^{n}}}}$ $S_{n} = \frac{e^{2} + 1}{e^{2} - 1} - \frac{1}{2^{n}} . \frac{e^{\frac{1}{2^{n - 1}}} + 1}{e^{\frac{1}{2^{n - 1}}} - 1}$
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