Using the integral test, how do you show whether $(1 + (1/x))^x$ diverges or converges?

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Answer 1 · 2015-12-23T21:30:40

it converge

By the way you can process like this

$(1+1/x)^x=e^(xln(1+1/x)$

$1/(x+1)<=ln(1+1/x)<=1/x$

$x/(x+1)<=xln(1+1/x)<=1$

$1/(1+1/x)<=xln(1+1/x)<=1$

Take the limit of $1/(1+1/x)$ at $-oo$ and $oo$ for both case it's 1

by the squeeze theorem you can say that $lim x-> oo$ $xln(1+1/x) = 1$
and $lim x-> -oo$ $xln(1+1/x) = 1$

So $(1+1/x)^x$ converge

Using the integral test, how do you show whether (1 + (1/x))^x (1 + (1/x))^x diverges or converges?