Infinite Series

  • Thread starter godtripp
  • Start date
  • #1
54
0
[tex]\sum\frac{1}{n(n+k)}[/tex] from n=1 to infinity

find that the series is convergent and find it's sum.

Now I'm a bit confused... I can show it's convergent with k=1
and I attempted the same thing with k by breaking this into partial fractions. But I'm given a harmonic series that is divergent minus another divergent series... how can this be convergent?
 

Answers and Replies

  • #2
22,129
3,297
Well, let's look at the series [tex]\sum{\frac{k}{n(n+k)}[/tex] instead. You are probably aware that you can split this in partial fractions:

[tex]\frac{k}{n(n+k)}=\frac{1}{n}-\frac{1}{n+k}[/tex]

Now it's not immediately clear what happens if you sum the above series. Try taking k=2 and write 10 terms of the above series. You will see that a lot of terms vanish. This will give you an idea for the general proof...
 
  • #3
54
0
Thank you micromass, I'll try expanding that out as soon as I get home. I'm sure it'll telescope out... there is one thing driving me nuts however.

[tex]\sum 1/n[/tex] is a harmonic series... which is divergent.

I'm not sure about [tex]\sum 1/(n+k)[/tex]

since series have the property that [tex]\sum (a-b)[/tex]=[tex]\sum a[/tex] - [tex]\sum b[/tex]

how is it that the difference between a divergent series and a convergent (or divergent series) results in a convergent series?
 
  • #4
22,129
3,297
Well, for one thing, the series [tex]\sum{\frac{1}{n+k}}[/tex] is divergent.

But that aside, you state the equality

[tex]\sum(a_n-b_n)=\sum a_n - \sum b_n [/tex]

This statement is INCORRECT. This is only correct is both sequences are convergent. Thus this equality is not applicable in this case.
 
  • #5
54
0
Thank you!
 

Related Threads on Infinite Series

  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
2
Views
840
  • Last Post
Replies
1
Views
708
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
5
Views
3K
  • Last Post
Replies
2
Views
755
  • Last Post
Replies
4
Views
973
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
7
Views
2K
Top