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not convinced - need to show monotonic increasing nature of function
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To be more rigourous you could have written that (1+1/1000)^1000 < e and then substitute using an inequality, instead of using ~
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There is no need of doing nothing. Its obvious that 1000^1001 is more than 1001^1000
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1001^1000 = (1+1/1000)^1000 × 1000 ^1000 < e × 1000^1000 < 1000 × 1000^1000 = 1000^1001
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The exponent is a give away. No elaborate proof. Sheer fundamental algebra. Bon Jrz!
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Takipledim ^^
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(1+1/n)^n<3 if n>3 n>(1+1/n)^n n^(n+1) > (n+1)^n 1000^1001 > 1001^1000 q.e.d
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That 1 of 1001 is worth an extra 000
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