# What is limit (sumation(k=1 to n) k!(k^2+k+1))/((n+1)(n+1)!) if n go to infinity?

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### 1 Answer

You need to evaluate the following limit, such that:

`lim_(n->oo) (sum_(k = 1)^n (k!(k^2 + k + 1)))/((n+1)(n+1)!)`

You may replace `(k + 1)^2 - k` for `k^2 + k + 1` such that:

`lim_(n->oo) (sum_(k = 1)^n (k!((k + 1)^2 - k)))/((n+1)(n+1)!)`

`lim_(n->oo) (sum_(k = 1)^n (k!(k + 1)^2 - k!k))/((n+1)(n+1)!)`

You may replace `(k+1)!(k+1)` for `k!(k + 1)^2` such that:

`lim_(n->oo) (sum_(k = 1)^n ((k+1)!(k+1) - k!k))/((n+1)(n+1)!)`

Assigning values to k yields:

`k = 1 => 2!*2 - 1!*1`

`k = 2 => 3!*3 - 2!*2`

`k = 3 => 4!*4 - 3!*3`

......

`k = n - 1 => n!*n - (n+1)!(n+1)`

`k = n => (n+1)!(n+1) - n!*n`

Performing the addition of terms yields:

`sum_(k = 1)^n (k+1)!(k+1) - k!k = 2!*2 - 1!*1 + 3!*3 - 2!*2 + 4!*4 - 3!*3 + ... + n!*n - (n+1)!(n+1) + (n+1)!(n+1) - n!*n`

Reducing duplicate members yields:

`sum_(k = 1)^n (k+1)!(k+1) - k!k = (n+1)!(n+1) - 1!*1`

Replacing `(n+1)!(n+1) - 1!*1` for `sum_(k = 1)^n (k+1)!(k+1) - k!k` yields:

`lim_(n->oo) (sum_(k = 1)^n (k!((k + 1)^2 - k)))/((n+1)(n+1)!) = lim_(n->oo) ((n+1)!(n+1) - 1!*1)/((n+1)(n+1)!)`

`lim_(n->oo) ((n+1)!(n+1) - 1!*1)/((n+1)(n+1)!) = lim_(n->oo) ((n+1)!(n+1))/((n+1)(n+1)!) - lim_(n->oo) 1/((n+1)(n+1)!)`

Reducing duplicate factors yields:

`lim_(n->oo) ((n+1)!(n+1) - 1!*1)/((n+1)(n+1)!) = lim_(n->oo) 1 - lim_(n->oo) 1/((n+1)(n+1)!)`

Since `lim_(n->oo) 1/((n+1)(n+1)!) -> 0` yields:

`lim_(n->oo) ((n+1)!(n+1) - 1!*1)/((n+1)(n+1)!) = 1 - 0`

`lim_(n->oo) ((n+1)!(n+1) - 1!*1)/((n+1)(n+1)!) = 1`

**Hence, evaluating the given limit, yields **`lim_(n->oo) (sum_(k = 1)^n (k!(k^2 + k + 1)))/((n+1)(n+1)!) = 1.`