- #1

- 172

- 0

infinity

{Sigma} 2/n(n+2)

n=1

so i used partial fractions and got:

{Sigma} [1/n + 1/(n+2)]

then i used telescoping form to get the nth partial sum...

Sn= (1/1 + 1/3) + (1/2 + 1/4) +...

then i got the nth partial sum to be = 1+1/(n+2)

so the series converges and its sum is 1?

Does that seem right to everyone?