Telescopic sum issues, cant get Sk

[yuming]

1. Homework Statement
Find a formula for the nth partial sum of the series and use it to find the series' sum if the series converges

Homework Equations

3/(1*2*3) + 3,/(2*3*4) + 3/(3*4*5) +...+ 3/n(n+1)(n+2)

The Attempt at a Solution

the first try, i tried using partial fraction which equals to A/(n) + B/(n+1) + C/(n+2). which made me get (3/2n - 3/n+1 + 3/2(n+2). Sn= (3/2 - 3/2 +3/6) + (3/4 - 1 + 3/8) + (3/6 - 3/4 + 3/10) + (3/8 - 3/5 + 3/12). i can't cancel it correctly. please help
i saw the correct working which the partial fraction suppose to be A/(n)(n+1) - B/(n+1)(n+2). didnt make sense to me because i thought in partial fraction we are suppose to split all the parts? and i tried that way and i would get A/(n)(n+1) + B/(n+1)(n+2) instead of A/(n)(n+1) - B/(n+1)(n+2). (couldnt get negative b) please help, have been trying to solve this for days.

 

[Ray Vickson]

1. Homework Statement
Find a formula for the nth partial sum of the series and use it to find the series' sum if the series converges

Homework Equations

3/(1*2*3) + 3,/(2*3*4) + 3/(3*4*5) +...+ 3/n(n+1)(n+2)

The Attempt at a Solution

the first try, i tried using partial fraction which equals to A/(n) + B/(n+1) + C/(n+2). which made me get (3/2n - 3/n+1 + 3/2(n+2). Sn= (3/2 - 3/2 +3/6) + (3/4 - 1 + 3/8) + (3/6 - 3/4 + 3/10) + (3/8 - 3/5 + 3/12). i can't cancel it correctly. please help
i saw the correct working which the partial fraction suppose to be A/(n)(n+1) - B/(n+1)(n+2). didnt make sense to me because i thought in partial fraction we are suppose to split all the parts? and i tried that way and i would get A/(n)(n+1) + B/(n+1)(n+2) instead of A/(n)(n+1) - B/(n+1)(n+2). (couldnt get negative b) please help, have been trying to solve this for days.

Suggestions: (i) forget the '3' in the numerator; you can always put it back after you have done the summation; (ii) write $\frac{1}{n(n+2)}$ in partial fractions, then multiply by $1/(n+1)$ afterward.

 

[Mark44]

1. Homework Statement
Find a formula for the nth partial sum of the series and use it to find the series' sum if the series converges

Homework Equations

3/(1*2*3) + 3,/(2*3*4) + 3/(3*4*5) +...+ 3/n(n+1)(n+2)

The Attempt at a Solution

the first try, i tried using partial fraction which equals to A/(n) + B/(n+1) + C/(n+2). which made me get (3/2n - 3/n+1 + 3/2(n+2). Sn= (3/2 - 3/2 +3/6) + (3/4 - 1 + 3/8) + (3/6 - 3/4 + 3/10) + (3/8 - 3/5 + 3/12). i can't cancel it correctly. please help

First off, you need more parentheses. When you write 3/2n, that means $\frac 3 2 n$, not $\frac3 {2n}$ as you intended. When you write 3/n + 1, that's even worse, as it means $\frac 3 n + 1$

i saw the correct working which the partial fraction suppose to be A/(n)(n+1) - B/(n+1)(n+2). didnt make sense to me because i thought in partial fraction we are suppose to split all the parts? and i tried that way and i would get A/(n)(n+1) + B/(n+1)(n+2) instead of A/(n)(n+1) - B/(n+1)(n+2). (couldnt get negative b) please help, have been trying to solve this for days.

You can split it however you want. I wouldn't have thought of this approach, but working it through, it makes sense.

Decompose by setting $\frac{1}{n(n + 1)(n + 2)} = \frac{A}{n(n + 1)} + \frac{B}{(n + 1)(n + 2)}$. After you find A and B, you will have a series that telescopes nicely.
Note: to simplify things you can take out a factor of 3 from all of the terms in the original problem.