# Thread: Determining absolute/conditional convergence, approximating sum

1. ## Determining absolute/conditional convergence, approximating sum

Determine if $\displaystyle \sum_{n=0}^{\infty} \frac {(-1)^{n+1}}{n^5+n+6}$ converges absolutely or conditionally. Approximate the sum within an accuracy of 0.01.

$\displaystyle \sum_{n=0}^{\infty} |\frac {(-1)^{n+1}}{n^5+n+6}|$
$\displaystyle \sum_{n=0}^{\infty} \frac {1}{n^5+n+6}$
Basic comparison test with $\displaystyle \sum \frac {1} {n^5}$.
$\displaystyle \sum \frac {1} {n^5}$ converges, so $\displaystyle \sum_{n=0}^{\infty} \frac {1}{n^5+n+6}$ converges as well. The series converges absolutely.

I'm having trouble understanding how to approximate. Do I set up random partial sums until it's less than 0.01? How do I know how how many terms I need for it to be less than 0.01? Is there a formula?

2. ## Re: Determining absolute/conditional convergence, approximating sum

Originally Posted by deezy
Determine if $\displaystyle \sum_{n=0}^{\infty} \frac {(-1)^{n+1}}{n^5+n+6}$ converges absolutely or conditionally. Approximate the sum within an accuracy of 0.01.

$\displaystyle \sum_{n=0}^{\infty} |\frac {(-1)^{n+1}}{n^5+n+6}|$
$\displaystyle \sum_{n=0}^{\infty} \frac {1}{n^5+n+6}$
Basic comparison test with $\displaystyle \sum \frac {1} {n^5}$.
$\displaystyle \sum \frac {1} {n^5}$ converges, so $\displaystyle \sum_{n=0}^{\infty} \frac {1}{n^5+n+6}$ converges as well. The series converges absolutely.

I'm having trouble understanding how to approximate. Do I set up random partial sums until it's less than 0.01? How do I know how how many terms I need for it to be less than 0.01? Is there a formula?
since the series is alternating , error < |first omitted term|

$\displaystyle Error < \frac{1}{100} \le \frac{1}{n^5+n+6}$

looks like $\displaystyle n = 3$ will get you there

so ...

$\displaystyle \sum_{n=0}^{\infty} \frac{(-1)^{n+1}}{n^5+n+6} \approx -\frac{1}{6} + \frac{1}{8} - \frac{1}{40}$ will be w/in the desired error bound