# Thread: [SOLVED] A Series with Partial Sums

1. ## [SOLVED] A Series with Partial Sums

Investigate the convergence of:

$\displaystyle \sum_{k=1}^{\infty}\,\,\frac{4}{k(k+2)}$.

So I broke it down into:

$\displaystyle \sum_{k=1}^{\infty}\,\,\frac{2}{k}-\frac{2}{k+2}$

I have no idea how to write this as the nth partial sum. I tried writing the first few terms:

$\displaystyle S_{3}=2-\frac{2}{3}+1-\frac{1}{2} + \frac{2}{3}-\frac{2}{5}$

I'm completely lost. Thanks!

2. Okay, so looking at it a bit more, and it looks like it'll be $\displaystyle S_{n} = 3 - \frac{2}{n+2}$

Is this correct?

A similar problem

$\displaystyle \sum_{k=1}^{\infty}\,\,\frac{9}{k(k+3)}$

3. Originally Posted by Pinkk
Okay, so looking at it a bit more, and it looks like it'll be $\displaystyle S_{n} = 3 - \frac{2}{n+2}$

Is this correct?
unfortunately not! the correct answer is: $\displaystyle S_n=3 - 2 \left(\frac{1}{n+1} + \frac{1}{n+2} \right).$ just find $\displaystyle S_n$ for a few values of $\displaystyle n$ and you'll see the pattern!

4. Ah, I think I get it. So using $\displaystyle S_{3}$ in the first problem (since this when terms begin to cancel each other out) you eventually arrive at:

$\displaystyle S_{3}= 3-\frac{2}{4}-\frac{2}{5}=3 - 2(\frac{1}{3+1}+\frac{1}{3+2})$ and then simply substitute $\displaystyle n=3$ in the denominators and I get what you got.

So for the second problem I posted, is this correct?

$\displaystyle S_{n}=\frac{11}{2} - 3(\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3})$ and thus leading to:

$\displaystyle \sum_{k=1}^{\infty}\,\,\frac{9}{k(k+3)} = \frac{11}{2}$

5. Originally Posted by Pinkk
Ah, I think I get it. So using $\displaystyle S_{3}$ in the first problem (since this when terms begin to cancel each other out) you eventually arrive at:

$\displaystyle S_{3}= 3-\frac{2}{4}-\frac{2}{5}=3 - 2(\frac{1}{3+1}+\frac{1}{3+2})$ and then simply substitute $\displaystyle n=3$ in the denominators and I get what you got.

So for the second problem I posted, is this correct?

$\displaystyle S_{n}=\frac{11}{2} - 3(\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3})$ and thus leading to:

$\displaystyle \sum_{k=1}^{\infty}\,\,\frac{9}{k(k+3)} = \frac{11}{2}$
correct! good job!

6. Thanks again. My professor is kinda rushing through this stuff since we're a bit behind schedule, not to mention we have a test on double and triple integration tomorrow, so my brain is on overload right now.