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{ln(n+2)/(n+2)} n=1 diff to find sequence?
use diff to find if this sequence is inc or dec
{ln(n+2)/(n+2)} n=1 ?? how to do this anyone?
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- Alfredo KrausLv 71 decade agoFavorite Answer
If y= ( ln (x) ) /x then dy/dx = (x (1/x) - ln(x) ) / x^2 = (1-ln(x)) / x^2. For n>=1, n+2>=3>e so
ln(n+2) > 1 and dy/dx will be negative, so the sequence is decreasing
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