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

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

$$\mathrm{Calculate}\:\mathrm{the}\:\mathrm{sum}\:\underset{\mathrm{k}\:=\:\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\frac{\mathrm{1}}{\:\sqrt{\mathrm{k}+\sqrt{\mathrm{k}^{\mathrm{2}} +\mathrm{1}}}}\:. \\ $$

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))).

$${let}\:{a}\:=\:{k}\:+\:\sqrt{{k}^{\mathrm{2}} −\mathrm{1}}\:{and}\:{b}\:=\:{k}−\sqrt{{k}^{\mathrm{2}} +\mathrm{1}}\: \\ $$$${then}\:{we}\:{have}\:\rightarrow\begin{cases}{{a}+{b}=\mathrm{2}{k}}\\{{ab}={k}^{\mathrm{2}} −\left(\sqrt{{k}^{\mathrm{2}} −\mathrm{1}}\right)^{\mathrm{2}} =\mathrm{1}}\end{cases} \\ $$$${so}\:{we}\:{get}\:\frac{\mathrm{1}}{\:\sqrt{{k}+\sqrt{{k}^{\mathrm{2}} +\mathrm{1}}}}\:=\:\frac{\mathrm{1}}{\:\sqrt{{a}}}\:=\:{b}. \\ $$$${On}\:{the}\:{other}\:{hand},\:{observe}\:{that} \\ $$$${b}\:=\:{k}−\sqrt{{k}^{\mathrm{2}} −\mathrm{1}}\:={k}−\sqrt{\left({k}+\mathrm{1}\right)\left({k}−\mathrm{1}\right)} \\ $$$$=\:\frac{{k}+\mathrm{1}+{k}−\mathrm{1}}{\mathrm{2}}−\sqrt{\left({k}+\mathrm{1}\right)\left({k}−\mathrm{1}\right)} \\ $$$$=\:\frac{{k}+\mathrm{1}−\mathrm{2}\sqrt{\left({k}+\mathrm{1}\right)\left({k}−\mathrm{1}\right)}+{k}−\mathrm{1}}{\mathrm{2}} \\ $$$$=\:\frac{\left(\sqrt{{k}+\mathrm{1}}−\sqrt{{k}−\mathrm{1}}\right)^{\mathrm{2}} }{\mathrm{2}} \\ $$$${hence}\:\sqrt{{b}}\:=\:\frac{\sqrt{{k}+\mathrm{1}}−\sqrt{{k}−\mathrm{1}}}{\:\sqrt{\mathrm{2}}} \\ $$$$\sqrt{{b}}\:=\:\frac{\sqrt{{k}+\mathrm{1}}−\sqrt{{k}}+\sqrt{{k}}−\sqrt{{k}−\mathrm{1}}}{\:\sqrt{\mathrm{2}}} \\ $$$${now}\:{the}\:{equation}\:{can}\:{be}\:{rewritte}\:{as}\: \\ $$$$\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{\:\sqrt{{k}+\sqrt{{k}^{\mathrm{2}} +\mathrm{1}}}}\:=\:\underset{{k}\:=\:\mathrm{1}} {\overset{{n}} {\sum}}\frac{\sqrt{{k}+\mathrm{1}}−\sqrt{{k}}+\sqrt{{k}}−\sqrt{{k}−\mathrm{1}}}{\:\sqrt{\mathrm{2}}} \\ $$$${by}\:{telescoping}\:{series}\:{gives} \\ $$$$=\:\frac{\sqrt{{n}+\mathrm{1}}−\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:+\:\frac{\sqrt{{n}}−\sqrt{\mathrm{0}}}{\:\sqrt{\mathrm{2}}} \\ $$$$=\:\frac{\sqrt{{n}+\mathrm{1}}+\sqrt{{n}}−\mathrm{1}}{\:\sqrt{\mathrm{2}}}.\: \\ $$

Commented by liberty last updated on 12/Nov/20

correct and great...

$$\mathrm{correct}\:\:\:\mathrm{and}\:\mathrm{great}... \\ $$

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