Calculate-k-1-n-1-e-1-k-1-

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

Calculate \dots

Ω = \underset{k = 1}{\sum^{n}} ⌊ \frac{1}{\sqrt[k]{e} - 1} ⌋ = ?

Answered by TonyCWX08 last updated on 18/Feb/24

w h e n k = 1

⌊ \frac{1}{e - 1} ⌋ = ⌊ 0.582 ⌋ = 0

w h e n k = 2

⌊ \frac{1}{\sqrt{e} - 1} ⌋ = ⌊ 1.541 ⌋ = 1

w h e n k = 3

⌊ \frac{1}{e^{\frac{1}{3}} - 1} ⌋ = ⌊ 2.528 ⌋ = 2

w h e n k = 4

⌊ \frac{1}{e^{\frac{1}{4}} - 1} ⌋ = ⌊ 3.521 ⌋ = 3

S o

\underset{k = 1}{\sum^{n}} ⌊ \frac{1}{\sqrt[k]{e} - 1} ⌋

= \underset{k = 1}{\sum^{n}} (k - 1)

= \frac{n^{2}}{2}

Answered by witcher3 last updated on 18/Feb/24

e^{\frac{1}{k}}

1 + x + \frac{x^{2}}{2} e ⩾ e^{x} > 1 + x

\Rightarrow \frac{1}{k} + \frac{e}{{2 k}^{2}} ⩾ e^{\frac{1}{k}} - 1 < \frac{1}{k}

\Rightarrow \frac{{2 k}^{2}}{2 k + e} ⩽ \frac{1}{e^{\frac{1}{k}} - 1} ⩽ k

\Rightarrow k - 1 < k - \frac{e}{2 k + e} ⩽ \frac{1}{e^{\frac{1}{k}} - 1} < k

\Rightarrow [\frac{1}{e^{\frac{1}{k}} - 1}] = k - 1

\underset{k = 1}{\sum^{n}} k - 1 = \frac{n (n - 1)}{2}

Commented by TonyCWX08 last updated on 19/Feb/24

t h e a n s w e r i s w r o n g .

\underset{k = 1}{\sum^{n}} k = \frac{n (n - 1)}{2}

B u t

\underset{k = 1}{\sum^{n}} k - 1 \neq \frac{n (n - 1)}{2}

i t s h o u l d b e e q u a l t o \frac{n^{2}}{2}

Commented by witcher3 last updated on 19/Feb/24

\frac{n^{2}}{2} \notin N for n = 2 k + 1

\underset{1}{\sum^{n}} k = \frac{n (n + 1)}{2}

\underset{k = 1}{\sum^{n}} (k + a) = \frac{n (n + 2 a + 1))}{2}

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

i t w a s m e a n t :

\underset{k = 1}{\sum^{n}} (k - 1) = \frac{n (n - 1)}{2}

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