# Thread: Error in alternating series

1. ## Error in alternating series

I really need someone to walk me through this step by step. I have no idea how to go about this.

Show that the series is convergent. How many terms of the series do we need to add in order to find the sum to the indicated accuracy?

$\displaystyle \sum^{\infty}_{n=1} \frac{(-1)^{n+1}}{7n^4}$

$\displaystyle (|error|) < .00005$

2. In an alternating series with the absolute value of the terms decreasing towards zero we have...

$\displaystyle \biggl|S_n-L\biggr|<a_{n+1}$

where L is the value of the entire sum, $\displaystyle S_n$ is the n$\displaystyle ^{th}$ partial sum and $\displaystyle a_{n+1}$ is the next term.

3. Originally Posted by mollymcf2009
I really need someone to walk me through this step by step. I have no idea how to go about this.

Show that the series is convergent. How many terms of the series do we need to add in order to find the sum to the indicated accuracy?

$\displaystyle \sum^{\infty}_{n=1} \frac{(-1)^{n+1}}{7n^4}$

$\displaystyle (|error|) < .00005$
note that the basic series is an alternating p-series, p > 1, so convergence is absolute.

error for an alternating series is less than the first omitted term ...

$\displaystyle \left|\frac{(-1)^n}{7n^4}\right| < .00005$

$\displaystyle \frac{1}{7n^4} < .00005$

$\displaystyle 7n^4 > \frac{1}{.00005} = 20000$

$\displaystyle 7 \cdot 7^4 = 7^5 = 16807$

$\displaystyle 7 \cdot 8^4 = 28672$

$\displaystyle n = 7$ should do it

4. ## Thank you!!!

Thanks Skeeter! That is EXACTLY what I was looking for!!