# Evaluate the limit of string [2^2+4^2+6^2+...+(2n-2)^2]/n^3 if the number of terms of string is infinite?

*print*Print*list*Cite

### 2 Answers

We have to find lim n-->inf.[2^2+4^2+6^2+...+(2n-2)^2]/n^3

[2^2+4^2+6^2+...+(2n-2)^2]/n^3

=> 2^2[1^2 + 2^2 + 3^2+...+(n - 1)^2]/n^3

=> 2^2((n - 1)*n*(2n - 1)/6)/n^3

lim n-->inf.[(2^2+4^2+6^2+...+(2n-2)^2)/n^3]

=> lim n-->inf.[2^2*(n - 1)*n*(2n - 1)/6*n^3]

=> lim n-->inf.[2^2*(n - 1)(2n - 1)/6*n^2]

=> lim n-->inf.[2^2*(2n^2 - 3n + 1)/6*n^2]

=> lim n-->inf.[(2^2/6)*(2n^2/n^2 - 3n/n^2 + 1/n^2)]

=> lim n-->inf.[(2^2/6)*(2 - 3/n + 1/n^2)]

if n--> inf. (1/n)--> 0

=> 2*4/6

=> 8/6

=> 4/3

**The required value of the limit is 4/3**

To evaluate this limit, we'll need the Cezaro-Stolz's theorem. This theorem states that if the limit of the ratio (un+1 - un)/(vn+1 - vn) exists, then the limit of the ratio un/vn existsÂ and it is equal to the previous limit.

We'll put the sum from numerator as un = 2^2+4^2+6^2+...+(2n-2)^2

un+1 =2^2+4^2+6^2+...+(2n-2)^2 + [2(n+1) - 2]^2

un+1 = 2^2+4^2+6^2+...+(2n-2)^2 + (2n)^2

un+1 - un = 2^2+4^2+6^2+...+(2n-2)^2 + (2n)^2 - 2^2 - 4^2 - 6^2-...- (2n-2)^2

We'll eliminate like terms and we'll get:

un+1 - un = (2n)^2

We'll put vn = n^3

vn+1 = (n+1)^3

vn+1 - vn = (n+1)^3 - n^3

We'll have to prove that the limit of the ratio (2n)^2/[(n+1)^3 - n^3] exists.

We'll expand the cube from denominator:

(n+1)^3 = n^3 + 3n^2 + 3n + 1

We'll subtract n^3:

[(n+1)^3 - n^3] = n^3 + 3n^2 + 3n + 1- n^3

[(n+1)^3 - n^3] = 3n^2 + 3n + 1

We'll re-write the limit:

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

We'll factorize by n^2:

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

lim 2/(3 + 3/n + 1/n^2) = 2/3, if n approaches to infinite

**Since the limit of the ratio (2n)^2/[(n+1)^3 - n^3] = 2/3, then the limit of the ratio [2^2+4^2+6^2+...+(2n-2)^2]/n^3 = 2/3.**