Telescoping series convergence question

[SYoungblood]

Homework Statement

[/B]
Hello, this problem is from a well-known calc text:

Σ(n=1 to ∞) 8/(n(n+2)

Homework Equations

[/B]
What I have here is decomposingg the problem into Σ(n=1 to ∞)(8/n -(8/n+2)

The Attempt at a Solution

I have the series sum as equaling (8/1-8/3) + (8/2-8/4) + (8/3-8/5) + (8/4-8/6) + … + (8/n-(8/n+2)

So, I have everything canceling but 8/1, 8/2, and 8/n+2. The last arm reaches 0 as n approaches ∞, so I have 8/1 + 8/2 = 12.

However, I think I am missing something here, the answer in the back of the book is not 12.

Thank you all for your help in advance.

SY

 

[SammyS]

Homework Statement

[/B]
Hello, this problem is from a well-known calc text:

Σ(n=1 to ∞) 8/(n(n+2))

Homework Equations

[/B]
What I have here is decomposing the problem into Σ(n=1 to ∞)(8/n -(8/(n+2))

The Attempt at a Solution

I have the series sum as equaling (8/1-8/3) + (8/2-8/4) + (8/3-8/5) + (8/4-8/6) + … + (8/n-(8/(n+2))

So, I have everything canceling but 8/1, 8/2, and 8/(n+2). The last arm reaches 0 as n approaches ∞, so I have 8/1 + 8/2 = 12.

However, I think I am missing something here, the answer in the back of the book is not 12.

Thank you all for your help in advance.

SY

Check your algebra.

$\displaystyle \ \frac{8}{n}-\frac{8}{n+2}=\frac{8n+16-8n}{n(n+2)} \$

$\displaystyle \ \quad \quad \ne\frac{8}{n(n+2)} \$​

Also, be sure to use an adequate amount of parentheses.