# Thread: converging serries

1. ## converging serries

question: find out if the following converges and if it does find what number it converges to: $\sum_{n=1}^{\infty} \frac{2}{n^2+4n+3}$ use a telescoping sum

work: I used partial fractions to get $\frac{2}{n^2+4n+3}=\frac{1}{n+1}-\frac{1}{n+3}$

then I used a telescoping sum and the first term and last term remained, which are $\frac{1}{2}+-\frac{1}{n+3}$ and then taking the limit n-->infinity I'm left with 1/2.

Is this right? When I graphed the function it appears to go to 1.

2. Originally Posted by superdude
question: find out if the following converges and if it does find what number it converges to: $\sum_{n=1}^{\infty} \frac{2}{n^2+4n+3}$ use a telescoping sum
work: I used partial fractions to get $\frac{2}{n^2+4n+3}=\frac{1}{n+1}-\frac{1}{n+3}$
then I used a telescoping sum and the first term and last term remained, which are $\frac{1}{2}+-\frac{1}{n+3}$ and then taking the limit n-->infinity I'm left with 1/2.
I get $\frac{1}{2}+\frac{1}{3}-\frac{1}{n+3}$ as a partial sum.

3. Originally Posted by Plato
I get $\frac{1}{2}+\frac{1}{3}-\frac{1}{n+3}$ as a partial sum.
If the first two terms survive then wouldn't the last two terms survive?

4. Originally Posted by Danny
If the first two terms survive then wouldn't the last two terms survive?
Yes that is correct.
But it does not change the sum. Does it?

5. Originally Posted by Plato
Yes that is correct.
But it does not change the sum. Does it?
No b/c they drop to zero as $n \to \infty$.

6. I see where I went wrong: I wrote out 3 terms and thought that all but the first and last number would cancel out in the sequence
e.g. $\frac{1}{2}+-\frac{-1}{n+3}$
how do I make sure that I don't make the same mistake in the future, I mean I can't write out infinitely many terms and I thought 3 would be sufficent to see the pattern. I mean isn't it pretty tricky when 1 term in the middle doesn't cancel out?

Originally Posted by Danny
If the first two terms survive then wouldn't the last two terms survive?
I am unclear as to what you mean. I thought the first and last term remained and the rest canceled out.
$
(\frac{1}{2}{\color{blue}-\frac{1}{4})+(\frac{1}{3}-\frac{1}{5})+\frac{1}{4}-\frac{1}{6}...+(\frac{1}{n+1}}-\frac{1}{n+3})$

what's in blue is what I thought canceled.

7. Originally Posted by superdude
I am unclear as to what you mean. I thought the first and last term remained and the rest canceled out.
$
(\frac{1}{2}{\color{blue}-\frac{1}{4})+(\frac{1}{3}-\frac{1}{5})+\frac{1}{4}-\frac{1}{6}...+(\frac{1}{n+1}}-\frac{1}{n+3})$

what's in blue is what I thought canceled.
There is nothing there to cancel out the $\frac{1}{3}$ term.

So the nth partial sum is $\frac{1}{2}+\frac{1}{3}-\frac{1}{n+2}-\frac{1}{n+3}$ as stated by Plato (and "corrected" by Danny)...