Finding the sum of the series

  • Thread starter trap101
  • Start date
  • #1
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?
 

Answers and Replies

  • #2
35,292
7,148
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?

Rewrite 1/(k(k + 3)) as two fractions, using partial fraction decomposition.
 
  • #3
DryRun
Gold Member
838
4
[tex]\sum^{\infty}_{k=1} \frac{1}{k(k+3)}[/tex]
First, express [itex]\frac{1}{k(k+3)}[/itex] as partial fractions.
 
  • #4
342
0
Darn. Have to go back and review that, but it makes sense to do that now, thanks.
 

Related Threads on Finding the sum of the series

  • Last Post
Replies
11
Views
1K
  • Last Post
Replies
21
Views
2K
  • Last Post
Replies
10
Views
3K
  • Last Post
Replies
20
Views
2K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
7
Views
3K
  • Last Post
Replies
5
Views
1K
Replies
10
Views
2K
Replies
2
Views
1K
  • Last Post
Replies
3
Views
1K
Top