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1.	lim1+2+3+....n/(1-n2)n→infinity
Answer» 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

lim1+2+3+....n/(1-n2)n→infinity

Answer»

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

Discussion

2.	lim1+2+3+....n/(1-n2)n→infinity
Answer» it should be (-1/2) \

Discussion