let k be natural number. defined s_k as the sum of the infinite series s_k =((k^2 −1)/k^0 ) + ((k^2 −1)/k^1 ) + ((k^2 −1)/k^2 ) +... find the value of Σ_(k=1) ^∞ [(s


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Question Number 192608 by York12 last updated on 22/May/23

$l e t k b e n a t u r a l n u m b e r . d e f i n e d s_{k} a s t h e$ $s u m o f t h e i n f i n i t e s e r i e s s_{k} = \frac{k^{2} - 1}{k^{0}} + \frac{k^{2} - 1}{k^{1}} + \frac{k^{2} - 1}{k^{2}} + . . .$ $f i n d t h e v a l u e o f \underset{k = 1}{\sum^{\infty}} [\frac{s_{k}}{2^{k - 1}}] .$
	Answered by witcher3 last updated on 24/May/23

	$S_{k} = \sum_{n ⩾ 0} \frac{(k^{2} - 1)}{k^{n}}, S_{1 = 0}$ $\forall k ⩾ 2 S_{k} = (k^{2} - 1) . \frac{1}{1 - \frac{1}{k}} = k (k + 1)$ $\sum_{k ⩾ 1} [\frac{S_{k}}{2^{k - 1}}] = \sum_{k ⩾ 2} [\frac{k (k + 1)}{2^{k - 1}}] = A$ $k (k + 1) < 2^{k - 1},$ $k = 7 . . . proof$ $6.7 < 2^{6} = 64$ $suppse \forall k ⩾ 7 k (k + 1) ⩽ 2^{k - 1}, (k + 1) (k + 2) ⩽ 2^{k}$ $we have 2 k (k + 1) ⩽ 2^{k}$ ${2 k}^{2} + 2 k = k^{2} + 3 k + 2 + k^{2} - k - 2$ $k^{2} - k - 2 = (k + 1) (k - 2) ⩾ 8.5 > 0$ $\Rightarrow true$ $\Rightarrow \forall k ⩾ 7 \frac{k (k + 1)}{2^{k - 1}} < 1 \Rightarrow [\frac{k (k + 1)}{2^{}}] = 0$ $A = \underset{k = 1}{\sum^{6}} [\frac{k (k + 1)}{2^{k - 1}}] + 0$ $= [2] + [\frac{6}{2}] + [\frac{12}{4}] + [\frac{20}{8}] + [\frac{30}{16}] + [\frac{42}{32}]$ $= 12$
	Answered by Rajpurohith last updated on 26/May/23

	$c l e a r l y s_{k} = (k^{2} - 1) \underset{i = 0}{\sum^{\infty}} (\frac{1}{k^{i}}) = (k^{2} - 1) . (\frac{1}{1 - \frac{1}{k}})$ $= k (k + 1)$ $s o \underset{k = 0}{\sum^{\infty}} (\frac{s_{k}}{2^{k - 1}}) = \underset{k = 0}{\sum^{\infty}} \frac{k (k + 1)}{2^{k - 1}}$
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