Partial Sum Formula of Telescoping Series: 1/2*(1+1/2-1/(n+1)-1/(n+2))

[chris4642]

Homework Statement

Find the formula of the partial sum of the series Ʃ1/[k(k+2)] k from 1 to infinity

Homework Equations

The Attempt at a Solution

Using partial fractions i rewrite the series as 1/2*Ʃ[1/k] - [1/(k+2)]

Then I start writing out the series from k=1 to 5.
1/2*[(1-1/3)+(1/2-1/4)+(1/3-1/5)+(1/4-1/6)+(1/5-1/7)+...+(1/n-1/(n+2))]

Everything cancels except for 1-1/6-1/7+(1/n-1/(n+2))

I think that S_n=1/2*(1+1/2-1/(n+2)-1/(n+2))

My professor in class solved it as 1/2*(1+1/2-1/(n+1)-1/(n+2))

I don't understand where the 1/n+1 term comes from. Is he right or am I, and why?

 

[LCKurtz]

Homework Statement

Find the formula of the partial sum of the series Ʃ1/[k(k+2)] k from 1 to infinity

Homework Equations

The Attempt at a Solution

Using partial fractions i rewrite the series as 1/2*Ʃ[1/k] - [1/(k+2)]

Then I start writing out the series from k=1 to 5.
1/2*[(1-1/3)+(1/2-1/4)+(1/3-1/5)+(1/4-1/6)+(1/5-1/7)+...+(1/n-1/(n+2))]

Apparently that is $\frac 1 n - \frac 1 {n+2}$. A little spacing would be good.

Everything cancels except for 1-1/6-1/7+(1/n-1/(n+2))

I think that S_n=1/2*(1+1/2-1/(n+2)-1/(n+2))

My professor in class solved it as 1/2*(1+1/2-1/(n+1)-1/(n+2))

I don't understand where the 1/n+1 term comes from. Is he right or am I, and why?

Your professor is correct. The last few terms are$$
...(\frac 1 {n-3}-\frac 1 {n-1})+(\frac 1 {n-2}-\frac 1 n ) + (\frac 1 {n-1} - \frac 1 {n+1}) +(\frac 1 n -\frac 1 {n+2})$$Only the second term of each of the last two parentheses remains.