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