As we know ;

1+2+3+.......+n = n(n+1)/2

So after substituting it's value

We get 

Lim(n-->infinity). [ n(n+1)]/[2(1-n²)]

That is  ==>     n/2(1-n) or -n/2(n-1)

And then take n common and cancel 'n' after cancellation we get 

lim(n-->infinity).  -1/2[1/[1-(1/n)] ]

As n--->infinity 1/n will tend to 0 

After applying limits we get 

Answer::  -1/2 or -0.5....

Thanks