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Question Number 204472 by mnjuly1970 last updated on 18/Feb/24

Calculate ... Ω=Σ_(k=1) ^n ⌊(( 1)/( (e)^(1/k) −1)) ⌋ =?

$$\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\mathrm{Calculate}\:... \\ $$$$\:\:\:\:\:\:\:\Omega=\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\:\lfloor\frac{\:\mathrm{1}}{\:\sqrt[{{k}}]{{e}}\:−\mathrm{1}}\:\rfloor\:=? \\ $$$$ \\ $$

Answered by TonyCWX08 last updated on 18/Feb/24

when k=1 ⌊(1/(e−1))⌋=⌊0.582⌋=0 when k = 2 ⌊(1/( (√e)−1))⌋=⌊1.541⌋=1 when k = 3 ⌊(1/(e^(1/3) −1))⌋=⌊2.528⌋=2 when k = 4 ⌊(1/(e^(1/4) −1))⌋=⌊3.521⌋=3 So Σ_(k=1) ^n ⌊(1/( (e)^(1/k) −1))⌋ =Σ_(k=1) ^n (k−1) =(n^2 /2)

$${when}\:{k}=\mathrm{1} \\ $$$$\lfloor\frac{\mathrm{1}}{{e}−\mathrm{1}}\rfloor=\lfloor\mathrm{0}.\mathrm{582}\rfloor=\mathrm{0} \\ $$$$ \\ $$$${when}\:{k}\:=\:\mathrm{2} \\ $$$$\lfloor\frac{\mathrm{1}}{\:\sqrt{{e}}−\mathrm{1}}\rfloor=\lfloor\mathrm{1}.\mathrm{541}\rfloor=\mathrm{1} \\ $$$$ \\ $$$${when}\:{k}\:=\:\mathrm{3} \\ $$$$\lfloor\frac{\mathrm{1}}{{e}^{\frac{\mathrm{1}}{\mathrm{3}}} −\mathrm{1}}\rfloor=\lfloor\mathrm{2}.\mathrm{528}\rfloor=\mathrm{2} \\ $$$$ \\ $$$${when}\:{k}\:=\:\mathrm{4} \\ $$$$\lfloor\frac{\mathrm{1}}{{e}^{\frac{\mathrm{1}}{\mathrm{4}}} −\mathrm{1}}\rfloor=\lfloor\mathrm{3}.\mathrm{521}\rfloor=\mathrm{3} \\ $$$$ \\ $$$${So} \\ $$$$\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\lfloor\frac{\mathrm{1}}{\:\sqrt[{{k}}]{{e}}−\mathrm{1}}\rfloor \\ $$$$=\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left({k}−\mathrm{1}\right) \\ $$$$=\frac{{n}^{\mathrm{2}} }{\mathrm{2}} \\ $$

Answered by witcher3 last updated on 18/Feb/24

e^(1/k) 1+x+(x^2 /2)e≥e^x >1+x ⇒(1/k)+(e/(2k^2 ))≥e^(1/k) −1<(1/k) ⇒((2k^2 )/(2k+e))≤(1/(e^(1/k) −1))≤k ⇒k−1<k−(e/(2k+e))≤(1/(e^(1/k) −1))<k ⇒[(1/(e^(1/k) −1))]=k−1 Σ_(k=1) ^n k−1=((n(n−1))/2)

$$\mathrm{e}^{\frac{\mathrm{1}}{\mathrm{k}}} \\ $$$$\mathrm{1}+\mathrm{x}+\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}\mathrm{e}\geqslant\mathrm{e}^{\mathrm{x}} >\mathrm{1}+\mathrm{x} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{k}}+\frac{\mathrm{e}}{\mathrm{2k}^{\mathrm{2}} }\geqslant\mathrm{e}^{\frac{\mathrm{1}}{\mathrm{k}}} −\mathrm{1}<\frac{\mathrm{1}}{\mathrm{k}} \\ $$$$\Rightarrow\frac{\mathrm{2k}^{\mathrm{2}} }{\mathrm{2k}+\mathrm{e}}\leqslant\frac{\mathrm{1}}{\mathrm{e}^{\frac{\mathrm{1}}{\mathrm{k}}} −\mathrm{1}}\leqslant\mathrm{k} \\ $$$$\Rightarrow\mathrm{k}−\mathrm{1}<\mathrm{k}−\frac{\mathrm{e}}{\mathrm{2k}+\mathrm{e}}\leqslant\frac{\mathrm{1}}{\mathrm{e}^{\frac{\mathrm{1}}{\mathrm{k}}} −\mathrm{1}}<\mathrm{k} \\ $$$$\Rightarrow\left[\frac{\mathrm{1}}{\mathrm{e}^{\frac{\mathrm{1}}{\mathrm{k}}} −\mathrm{1}}\right]=\mathrm{k}−\mathrm{1} \\ $$$$\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\mathrm{k}−\mathrm{1}=\frac{\mathrm{n}\left(\mathrm{n}−\mathrm{1}\right)}{\mathrm{2}} \\ $$

Commented by TonyCWX08 last updated on 19/Feb/24

the answer is wrong. Σ_(k=1) ^n k = ((n(n−1))/2) But Σ_(k=1) ^n k−1 ≠ ((n(n−1))/2) it should be equal to (n^2 /2)

$$ \\ $$$${the}\:{answer}\:{is}\:{wrong}. \\ $$$$\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}\:=\:\frac{{n}\left({n}−\mathrm{1}\right)}{\mathrm{2}} \\ $$$${But} \\ $$$$\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}−\mathrm{1}\:\neq\:\frac{{n}\left({n}−\mathrm{1}\right)}{\mathrm{2}} \\ $$$${it}\:{should}\:{be}\:{equal}\:{to}\:\frac{{n}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$ \\ $$

Commented by witcher3 last updated on 19/Feb/24

(n^2 /2)∉N for n=2k+1 Σ_1 ^n k=((n(n+1))/2) Σ_(k=1) ^n (k+a)=((n(n+2a+1)))/2)

$$\frac{\mathrm{n}^{\mathrm{2}} }{\mathrm{2}}\notin\mathbb{N}\:\:\mathrm{for}\:\mathrm{n}=\mathrm{2k}+\mathrm{1} \\ $$$$\underset{\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\mathrm{k}=\frac{\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)}{\mathrm{2}} \\ $$$$\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\left(\mathrm{k}+\mathrm{a}\right)=\frac{\left.\mathrm{n}\left(\mathrm{n}+\mathrm{2a}+\mathrm{1}\right)\right)}{\mathrm{2}} \\ $$

Commented by mr W last updated on 19/Feb/24

it was meant: Σ_(k=1) ^n (k−1)=((n(n−1))/2)

$${it}\:{was}\:{meant}: \\ $$$$\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\left(\mathrm{k}−\mathrm{1}\right)=\frac{\mathrm{n}\left(\mathrm{n}−\mathrm{1}\right)}{\mathrm{2}} \\ $$

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