# evaluate the limit of the series an=ln(n^2+n+1)/n

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The given series is convergent if it has a finite limit. We'll calculate the limit of this series forcing n^2 factor within the logarithm function.

lim an = lim ln[n^2*(1 + 1/n + 1/n^2)]/n

n->`oo` n->`oo`

We'll use the product property of logarithms:

lim an = lim [ln(n^2) + ln(1 + 1/n + 1/n^2)]/n

n->` oo` n->`oo`

We'll use the power property of logarithms:

lim an = lim 2*(1/n)ln n + lim ln(1 + 1/n + 1/n^2)]/n

n->`oo` n->`oo` n->`oo`

lim an = 2lim ln (n)^(1/n) + 0

n->`oo`

But lim ln (n)^(1/n) = 1

lim an = 2

n -> `oo`

**The limit of the given series, if n is approaching to `oo` , is lim an = 2.**