Calculate the sum Σ_(k = 1) ^n (1/( (√(k+(√(k^2 +1)))))) .


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Question Number 121950 by liberty last updated on 12/Nov/20

$Calculate the sum \underset{k = 1}{\sum^{n}} \frac{1}{\sqrt{k + \sqrt{k^{2} + 1}}} .$
	Answered by bobhans last updated on 12/Nov/20
	$let a = k + (√(k^2 −1)) and b = k−(√(k^2 +1)) then we have → { ((a+b=2k)),((ab=k^2 −((√(k^2 −1)))^2 =1)) :} so we get (1/( (√(k+(√(k^2 +1)))))) = (1/( (√a))) = b. On the other hand, observe that b = k−(√(k^2 −1)) =k−(√((k+1)(k−1))) = ((k+1+k−1)/2)−(√((k+1)(k−1))) = ((k+1−2(√((k+1)(k−1)))+k−1)/2) = ((((√(k+1))−(√(k−1)))^2 )/2) hence (√b) = (((√(k+1))−(√(k−1)))/( (√2))) (√b) = (((√(k+1))−(√k)+(√k)−(√(k−1)))/( (√2))) now the equation can be rewritte as Σ_(k=1) ^n (1/( (√(k+(√(k^2 +1)))))) = Σ_(k = 1) ^n (((√(k+1))−(√k)+(√k)−(√(k−1)))/( (√2))) by telescoping series gives = (((√(n+1))−1)/( (√2))) + (((√n)−(√0))/( (√2))) = (((√(n+1))+(√n)−1)/( (√2))).$
	$l e t a = k + \sqrt{k^{2} - 1} a n d b = k - \sqrt{k^{2} + 1}$ $t h e n w e h a v e \to {\begin{cases} a + b = 2 k \\ a b = k^{2} - {(\sqrt{k^{2} - 1})}^{2} = 1 \end{cases}$ $s o w e g e t \frac{1}{\sqrt{k + \sqrt{k^{2} + 1}}} = \frac{1}{\sqrt{a}} = b .$ $O n t h e o t h e r h a n d, o b s e r v e t h a t$ $b = k - \sqrt{k^{2} - 1} = k - \sqrt{(k + 1) (k - 1)}$ $= \frac{k + 1 + k - 1}{2} - \sqrt{(k + 1) (k - 1)}$ $= \frac{k + 1 - 2 \sqrt{(k + 1) (k - 1)} + k - 1}{2}$ $= \frac{{(\sqrt{k + 1} - \sqrt{k - 1})}^{2}}{2}$ $h e n c e \sqrt{b} = \frac{\sqrt{k + 1} - \sqrt{k - 1}}{\sqrt{2}}$ $\sqrt{b} = \frac{\sqrt{k + 1} - \sqrt{k} + \sqrt{k} - \sqrt{k - 1}}{\sqrt{2}}$ $n o w t h e e q u a t i o n c a n b e r e w r i t t e a s$ $\underset{k = 1}{\sum^{n}} \frac{1}{\sqrt{k + \sqrt{k^{2} + 1}}} = \underset{k = 1}{\sum^{n}} \frac{\sqrt{k + 1} - \sqrt{k} + \sqrt{k} - \sqrt{k - 1}}{\sqrt{2}}$ $b y t e l e s c o p i n g s e r i e s g i v e s$ $= \frac{\sqrt{n + 1} - 1}{\sqrt{2}} + \frac{\sqrt{n} - \sqrt{0}}{\sqrt{2}}$ $= \frac{\sqrt{n + 1} + \sqrt{n} - 1}{\sqrt{2}} .$
	Commented by liberty last updated on 12/Nov/20

	$correct and great . . .$
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