Find the smallest number of terms needed so that the error of the partial sum is less than $10^{-4}$
$$ \sum_{n=1}^\infty\frac{200(-1)^n}{75n^{0.89}+58} $$
Having trouble trying to even do this. If anyone has an example of can lead me to the final answer that would be great.
On
This is a standard type of question where one wants to discover how many terms of an alternating series one must add to get an answer accurately to a certain amount of precision.
For an alternating series the error is less than the absolute value of the term following the last term added.
So you are looking for the smallest value of $n$ for which
$$ \left\lvert\dfrac{200}{75(n+1)^{0.89}+58} \right\rvert<\dfrac{1}{10,000} $$
$$ 75(n+1)^{0.89}+58>2,000,000 $$
$$ (n+1)^{0.89}>\dfrac{1999942}{75}$$
$$ 0.89\log(n+1)>\log\left( \dfrac{1999942}{75}\right) $$
$$ \log(n+1)> \dfrac{1}{0.89}\log\left( \dfrac{1999942}{75}\right)$$
$$ \log(n+1)>4.972984424 $$
$$ n+1>93968.96084 $$
$$ n>93967.96084 $$
Therefore one must sum $93,968$ terms for that amount of precision.
The error of the partial sum of the first $n$ terms of an alternating series is $E\leq| a_{n+1} |$. In your case, you are trying to solve $10^{-4}\geq\frac{200}{75(n+1)^{0.89}+58}$ for $n$. Just treat this like a normal inequality, until you get something like $n\geq k$, where $k$ is some number. Then, you round $k$ up to get your final result, $n=k$.