Telescoping Series: Finding the Sum of 1/k(k+3)

[trap101]

So the question is find the sum of the series:


$\sum$ 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?

 

[Mark44]

So the question is find the sum of the series:


$\sum$ 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?

Rewrite 1/(k(k + 3)) as two fractions, using partial fraction decomposition.

 

[DryRun]

$$\sum^{\infty}_{k=1} \frac{1}{k(k+3)}$$
First, express $\frac{1}{k(k+3)}$ as partial fractions.

 

[trap101]

Darn. Have to go back and review that, but it makes sense to do that now, thanks.