Sigma (n=1 to infinity) [(1/(n+1))-(1/(n+2))]
Can anyone help me prove whether this is convergent or divergent?
I know I'm supposed to use telescoping series test, but am unsure how.
Sum from n=1 to n=N giving
[1/2 - 1/3]+[ 1/3-1/4]+[1/4-1/5]+....+[1/N - 1/(N+1)]+[1/(N+1)-1/(N+2)]
removing the brackets gives
=1/2-1/3+1/3-1/4+1/4-1/5+....+1/N - 1/(N+1)+1/(N+1)-1/(N+2)
and terms cancel to give
1/2 - 1/(N+2)
And as N->∞ this gives 1/2
so the series converges to 1/2
Comments
Sum from n=1 to n=N giving
[1/2 - 1/3]+[ 1/3-1/4]+[1/4-1/5]+....+[1/N - 1/(N+1)]+[1/(N+1)-1/(N+2)]
removing the brackets gives
=1/2-1/3+1/3-1/4+1/4-1/5+....+1/N - 1/(N+1)+1/(N+1)-1/(N+2)
and terms cancel to give
1/2 - 1/(N+2)
And as N->∞ this gives 1/2
so the series converges to 1/2