lim n^2/(1+2+3+...+n) n--> inf

We know that 1+2+3+...+n = n(n+1)/2

==> lim 2n^2/n(n+1) = 2lim n^2/ lim n^2+n

Let us divide by the highest power n^2

==> 2lim (n^2/n^2) / lim n^2(1+1/n)

Reduce similar:

==> 2 lim(1)/lim (1+1/n) when n--> inf

==> 2/1+0 = 2

To evaluate the limit of the rational function, when n tends to +inf.,we'll factorize both, numerator and denominator.

In this case, first, we'll write the sum of the denominator:

1+2+...+n = n*(n+1)/2

We'll substitute the denominator, by the result of the sum, we'll factorize by the highest power of n, which in this case is n^2.

We'll have:

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

lim n^2/lim (1+2+3+ ... +n) = lim 2*n^2/lim n*(n+1)

We'll open the brackets from the denominator:

lim 2*n^2/lim n^2(1+1/n)

We'll divide by n^2:

**lim 2/lim (1+1/n) = 2/(1+0)=2**

To find Lt n^2/(1+2+...n) as x--> infinity.

Solution:

The denominator is the sum of first n natural 1+2+3+.+n = n(n+1)/2.

So we have to find Lt n^2/(n(n+1)/2] = Lt 2n^2/(n^2+n) as n--> infinity. Since both numerator and denominator approach ifinity which is an indeterminate form, we use L'Hospital s rule of differentiating numerator and denominator and take the limit.

Limit of the given expression = (2n^2)'/(n^2+n)' = Lt 4n /(2n+n). Still in inf/inf form. Againg applying L'Hospoital's rule

limit of the expression = (4n)'/(2n+1)' = 4/2 = 2