# Use Stolz-Cesaro theorem for proving the convergence of the sequence if an=(182+2*3+..+n*(n+1))/n^3.

*print*Print*list*Cite

I'll solve this problem, considering that the term "182"i s 1*2 (since the "*" symbol and the digit 8 are on the same key and the rest of the terms are written accordingly).

The Stolz Cesaro theorem states that if there are 2 sequences, an and bn, where the sequence bn is unbounded and increasing and if the limit of the fraction ( a(n+1) - an)/(b(n+1) - bn) = l is finite then, the limit of the fraction an/bn = l, too.

Let an = 1*2 + 2*3 + ... + n*(n+1)

bn = n^3

We notice that bn is unbounded and increasing.

a(n+1) - an = 1*2 + 2*3 + ... + n*(n+1) + (n+1)*(n+2) - 1*2 - 2*3 - ... - n*(n+1)

We'll eliminate like terms:

a(n+1) - an = (n+1)*(n+2)

b(n+1) - bn = (n+1)^3 - n^3

b(n+1) - bn = n^3 + 3n^2 + 3n + 1 - n^3

We'll eliminate like terms:

b(n+1) - bn = 3n^2 + 3n + 1

We'll evaluate the limit:

lim (a(n+1) - an)/(b(n+1) - bn) = lim (n+1)*(n+2)/(3n^2 + 3n + 1), if n approaches to infinity.

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

Since the orders of polynomials from numerator and denominator are the same, then the limit is the ratio of leading coefficients:

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

**As we can see, the limit exists and it is finite, therefore the limit of the sequence [1*2 + 2*3 + ... + n*(n+1)]/n^3 = 1/3.**