- #1
trap101
- 342
- 0
So the question is find the sum of the series:
[itex]\sum[/itex] from k = 1 to ∞ of 1 / k(k+3),
now the solution they provided was:
= 1/3 [ (1 - 1/4) + (1/2 - 1/5) + ...+ (1/n - 1/(n+3) ]
=1/3 [ 1+ 1/2+1/3 -1/(n+1) - 1/(n+2) - 1/(n+3)]
--> 11/18
I'm stuck on how they were able to show the sum telescoping and why they were able to factor out the 1/3. Also how are you suppose to solve these sorts of questions?
[itex]\sum[/itex] from k = 1 to ∞ of 1 / k(k+3),
now the solution they provided was:
= 1/3 [ (1 - 1/4) + (1/2 - 1/5) + ...+ (1/n - 1/(n+3) ]
=1/3 [ 1+ 1/2+1/3 -1/(n+1) - 1/(n+2) - 1/(n+3)]
--> 11/18
I'm stuck on how they were able to show the sum telescoping and why they were able to factor out the 1/3. Also how are you suppose to solve these sorts of questions?