Calculate the limit (1+2+...+n)/(n^2+3n+1) n --> infinite

We'll substitute infinite into the given
expression:

 (1+2+...+inf.)/inf. =
inf./inf.

As we can see, this is a case of
indeterminacy

We notice that the numerator is the sum of
the first n terms of an arithmetic series and the formula for the sum
is:

S = (1+n)*n/2

We'll
substitute the numerator by the expression of the
sum:

limit
[(1+n)*n/2*(n^2+3n+1)]

We'll remove the brackets and we'll
draw out the constant value 1/2.

(1/2) * lim [(n^2 +
n)/(n^2+3n+1)]

We notice that the numerator and denominator
are polynomials that have the same degree, so the limit is the ratio of the coefficients
of the terms that have the highest degree.

In this case,
the ratio of the coefficients of n^2, both numerator and denominator, are
1/1.

The result of the limit
is:

limit (1+2+...+n)/(n^2+3n+1) =
1/2