# What is limit of series 1+2+3+4+5+......+n/n^2+3n+5?

You need to evaluate the limit of the given finite series such that:

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

You should substitute `n(n+1)/2`  for the sum of natural numbers `1+2+3+...+n`  such that:

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

Reducing by `n^2`  yields:

`lim_(n->oo) (n^2(1 + 1/n))/(n^2(1+3/n+5/n^2)) = lim_(n->oo) (1 + 1/n)/(2+6/n+10/n^2)`

Since the limits...

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You need to evaluate the limit of the given finite series such that:

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

You should substitute `n(n+1)/2`  for the sum of natural numbers `1+2+3+...+n`  such that:

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

Reducing by `n^2`  yields:

`lim_(n->oo) (n^2(1 + 1/n))/(n^2(1+3/n+5/n^2)) = lim_(n->oo) (1 + 1/n)/(2+6/n+10/n^2)`

Since the limits `lim_(n->oo) 1/n = lim_(n->oo) 6/n = lim_(n->oo) 10/(n^2) = 0`  yields:

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

Hence, evaluating the limit of the finite series yields `lim_(n->oo) (1+2+3+...+n)/(n^2+3n+5) = 1/2.`

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