Evaulate the Sum

1. Oct 24, 2012

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?

2. Oct 24, 2012

Dick

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.

3. Oct 24, 2012

BigJon

When i get to terms that cancel out, what do i do to solve thats where im confused. my teacher never explained exactly what to do

4. Oct 24, 2012

Zondrina

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

Last edited: Oct 24, 2012
5. Oct 24, 2012

BigJon

Yes so i just sum all those multiply by (1/8) and thats it? Im just trying to make sense of my teachers notes.

6. Oct 24, 2012

Dick

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

7. Oct 25, 2012

Ray Vickson

If $$S(N) = \sum_{n=1}^N \left[\frac{1}{n} - \frac{1}{n+8} \right] = \sum_{n=1}^N \frac{1}{n} - \sum_{n=1}^N \frac{1}{n+8},$$ 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

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