# How calculate limit of series 1*2+2*3+----+n(n+1)/1*3+2*5+----+n(2n++1)?

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### 1 Answer

You need to evaluate the sum to numerator such that:

`1*2 + 2*3 + ... + n(n+1) = sum_(k=1)^n k(k+1)`

`1*2 + 2*3 + ... + n(n+1) = sum_(k=1)^n (k^2 + k)`

`sum_(k=1)^n k^2 = 1^2 + 2^2 + .. + n^2 = (n(n+1)(2n+1))/6`

`sum_(k=1)^n k = 1 + 2 + .. + n = (n(n+1))/2`

`sum_(k=1)^n k(k+1) = (n(n+1)(2n+1))/6 + (n(n+1))/2`

Factoring out `n(n+1)` yields:

`sum_(k=1)^n k(k+1) = (n(n+1)(2n+1+3))/6`

`sum_(k=1)^n k(k+1) = (2n(n+1)(n+2))/6`

You need to evaluate the sum to denominator such that:

`1*3+2*5+...+n(2n+1) = sum_(k=1)^n k(2k+1)`

`sum_(k=1)^n k(2k+1) = sum_(k=1)^n (2k^2 + k)`

`2sum_(k=1)^n k^2 = 1^2 + 2^2 + .. + n^2 = 2(n(n+1)(2n+1))/6`

`2sum_(k=1)^n k^2 = (n(n+1)(2n+1))/3`

`sum_(k=1)^n k(2k+1) = (n(n+1)(2n+1))/3 + (n(n+1))/2`

Factoring out `n(n+1)` yields:

`sum_(k=1)^n k(2k+1) = (n(n+1)(4n+2+3))/6`

`sum_(k=1)^n k(2k+1) = (n(n+1)(4n+5))/6`

You need to evaluate the limit such that:

`lim_(n->oo)(1*2 + 2*3 + ... + n(n+1))/(1*3+2*5+...+n(2n+1)) = lim_(n->oo) ((2n(n+1)(n+2))/6)/((n(n+1)(4n+5))/6)`

`lim_(n->oo)(1*2 + 2*3 + ... + n(n+1))/(1*3+2*5+...+n(2n+1)) = lim_(n->oo) ((2n(n+1)(n+2)))/((n(n+1)(4n+5)))`

`lim_(n->oo)(1*2 + 2*3 + ... + n(n+1))/(1*3+2*5+...+n(2n+1)) = 2/4`

`lim_(n->oo)(1*2 + 2*3 + ... + n(n+1))/(1*3+2*5+...+n(2n+1)) = 1/2`

**Hence, evaluating the given limit yields `lim_(n->oo)(1*2 + 2*3 + ... + n(n+1))/(1*3+2*5+...+n(2n+1)) = 1/2` .**