# Thread: Approximating Sum of a Series

1. ## Approximating Sum of a Series

well I know this series converges absolutely after testing by comparison

Now I need to approximate the sum with an error less than .01

How do I know how far to go?
Thanks

2. ## Re: Approximating Sum of a Series

Originally Posted by minneola24
well I know this series converges absolutely after testing by comparison.
First it does not converge absolutely because $\sum\limits_n {\frac{{\sqrt n }}{{2n + 1}}}$ does not converge. WHY?

But it does converge conditionally using the alternating series test.

Look-up that test and that will be an approximating rule.

3. ## Re: Approximating Sum of a Series

Since that is an alternating series, you know that the "n+1" sum lies between the "n-1" sum and "n" sum. One of the things that means is that the infinite sum lies between any two consecutive terms. If two terms have a difference of less than .01, either of them is within .01 of the infinite sum.

4. ## Re: Approximating Sum of a Series

Ok I did the comparison test and your right I got conditionally convergent.

But I'm still confused how to calculus the sum with an error less than .01

Can I do trial and error on this?

5. ## Re: Approximating Sum of a Series

Originally Posted by minneola24
Ok I did the comparison test and your right I got conditionally convergent.
But I'm still confused how to calculus the sum with an error less than .01
Can I do trial and error on this?
If $a_n>0$ and $\sum\limits_{n = 1}^\infty {{{\left( { - 1} \right)}^n}{a_n}} = S$ and $S_n=\sum\limits_{n = 1}^\infty {{{\left( { - 1} \right)}^n}{a_n}}$ is a partial sum then $\left|S_n-S\right|.

So make $\frac{\sqrt{n+1}}{2n+3}<0.01$

6. ## Re: Approximating Sum of a Series

Thanks that worked perfect and I got the same answer as the book!

Now a new question how do I deal with factorial a? Such as this question

7. ## Re: Approximating Sum of a Series

New questions belong in new threads.

-Dan