How Does the Series Sum of 1/(n(n+8)) Converge?

[BigJon]

∞
Ʃ 1/(n(n+8))
n=1

So i used partial fractions and got (1/8)/(n) - (1/8)/(n+8)

From there i pulled out the 1/8 so now my equation is


∞
(1/8) Ʃ (1/n)-(1/(n+8))
n=1

So from here do i just start doing like s1= (1/8)(1-1/9), s2=(1/8)(1/2-1/10)

to find a value it converges to?

 

[Dick]

∞
Ʃ 1/(n(n+8))
n=1

So i used partial fractions and got (1/8)/(n) - (1/8)/(n+8)

From there i pulled out the 1/8 so now my equation is


∞
(1/8) Ʃ (1/n)-(1/(n+8))
n=1

So from here do i just start doing like s1= (1/8)(1-1/9), s2=(1/8)(1/2-1/10)

to find a value it converges to?

The sum of 1/n=1+1/2+1/3+1/4+...
The sum of 1/(n+8)=1/8+1/9+1/10+1/11+...
If you think about the difference, a LOT of terms cancel.

 

[BigJon]

When i get to terms that cancel out, what do i do to solve that's where I am confused. my teacher never explained exactly what to do

 

[STEMucator]

Notice that the terms 1 + 1/2 + 1/3 + ... stick around while the others get canceled?

 

[BigJon]

Yes so i just sum all those multiply by (1/8) and that's it? I am just trying to make sense of my teachers notes.

 

[Dick]

Yes so i just sum all those multiply by (1/8) and that's it? I am just trying to make sense of my teachers notes.

Yes, just sum the terms that don't cancel and divide by 8.

 

[Ray Vickson]

∞
Ʃ 1/(n(n+8))
n=1

So i used partial fractions and got (1/8)/(n) - (1/8)/(n+8)

From there i pulled out the 1/8 so now my equation is


∞
(1/8) Ʃ (1/n)-(1/(n+8))
n=1

So from here do i just start doing like s1= (1/8)(1-1/9), s2=(1/8)(1/2-1/10)

to find a value it converges to?

If S(N) = \sum_{n=1}^N \left[\frac{1}{n} - \frac{1}{n+8} \right]<br /> = \sum_{n=1}^N \frac{1}{n} - \sum_{n=1}^N \frac{1}{n+8}, try writing out a few of the S(N) values for N larger than 8, say for N = 10, 11, 12, etc. That will give you a feel for what is happening, and then the final solution will be easier to see.

RGV