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Question Number 146164 by mnjuly1970 last updated on 11/Jul/21

$c a l u l a t e ::$ $S : = \underset{n = 1}{\sum^{\infty}} \frac{H_{\frac{n}{2}}}{2^{n}} = ?$ $. . . . . . . m . n .$
	Answered by qaz last updated on 12/Jul/21
	$Σ_(n=1) ^∞ (1/2^n )∫_0 ^1 ((1−x^(n/2) )/(1−x))dx =∫_0 ^1 (1/(1−x))Σ_(n=1) ^∞ [(1/2^n )−(((√x)/2))^n ]dx =∫_0 ^1 (1/(1−x))(1−(((√x)/2)/(1−((√x)/2))))dx =2∫_0 ^1 (dx/((1+(√x))(2−(√x)))) =4∫_0 ^1 ((xdx)/((1+x)(2−x))) =(4/3)∫_0 ^1 ((2/(2−x))−(1/(1+x)))dx =(4/3){−2ln(2−x)−ln(1+x)}_0 ^1 =(4/3)ln2$
	$\underset{n = 1}{\sum^{\infty}} \frac{1}{2^{n}} \int_{0}^{1} \frac{1 - x^{n / 2}}{1 - x} dx$ $= \int_{0}^{1} \frac{1}{1 - x} \underset{n = 1}{\sum^{\infty}} [\frac{1}{2^{n}} - {(\frac{\sqrt{x}}{2})}^{n}] dx$ $= \int_{0}^{1} \frac{1}{1 - x} (1 - \frac{\frac{\sqrt{x}}{2}}{1 - \frac{\sqrt{x}}{2}}) dx$ $= 2 \int_{0}^{1} \frac{dx}{(1 + \sqrt{x}) (2 - \sqrt{x})}$ $= 4 \int_{0}^{1} \frac{xdx}{(1 + x) (2 - x)}$ $= \frac{4}{3} \int_{0}^{1} (\frac{2}{2 - x} - \frac{1}{1 + x}) dx$ $= \frac{4}{3} {- 2 \ln (2 - x) - \ln (1 + x)}_{0}^{1}$ $= \frac{4}{3} \ln 2$
	Commented by mnjuly1970 last updated on 12/Jul/21

	$t h x m r q a z . . . n i c e s o l u t i o n . . .$
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