Find the sum of the limit when Lim x => infinite [(1^1/x + 2^ 1/x..............2016^1/x)/2016]^2016x
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it form 1^(infinity)
now,
e^log [(1^1/x +2^1/x +..... 2016^1/x-2016)/2016]^2016x
e^2016xlog [(1^1/x +2^1/x +.....2016^1/x-2016)/2016]
e^2016log [(1^1/x+2^1/x+...2016^1/x-2016)/2016(1/x)]
e^log [{(1^1/x-1)/1/x} +{(2^1/x-1)/1/x}+... {(2016^1/x-1)/1/x}]
e^log [log1+log2+log3+log4+.....log2016]
e^log [log1 x 2 x 3 x 4 x ,,,.2016]
e^log (log2016!)
log (2016!)
now,
e^log [(1^1/x +2^1/x +..... 2016^1/x-2016)/2016]^2016x
e^2016xlog [(1^1/x +2^1/x +.....2016^1/x-2016)/2016]
e^2016log [(1^1/x+2^1/x+...2016^1/x-2016)/2016(1/x)]
e^log [{(1^1/x-1)/1/x} +{(2^1/x-1)/1/x}+... {(2016^1/x-1)/1/x}]
e^log [log1+log2+log3+log4+.....log2016]
e^log [log1 x 2 x 3 x 4 x ,,,.2016]
e^log (log2016!)
log (2016!)
enrique:
U r good
is answer correct
Yup
But I know a short trick
U did it with long method
what that
I join your chat please given trick
I will tell u in inbox because I don't want to tell it in comment box
Every one need points that's why
okay,
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