# Alt Series, Partial Sum

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.

• Are you talking about the error of the partial sum approximation or the sum itself being less than or equal to $10^{-4}$? – AlgorithmsX Nov 25 '16 at 2:23
• @AlgorithmsX The answer is n = ?, I am confused on this as well. – Robert Miller Nov 25 '16 at 2:25
• If you are looking for the error, then it's when $\frac{200}{75n^{0.89}+58}\leq10^{-4}$ by the rules of Alternating Series. – AlgorithmsX Nov 25 '16 at 2:27
• @AlgorithmsX Yes, I understand this far but don't know how to proceed afterwards.. – Robert Miller Nov 25 '16 at 2:29
• @AlgorithmsX: Actually, according to the Alternating Series Test, it's when the next term is less than the desired accuracy, i.e. when $\frac{200}{75(n+1)^{0.89}+58}\le10^{-4}$. What you said would provide an approximation within the desired accuracy, but not with the least $n$ (your $n$ would be the next after the least one). – zipirovich Nov 25 '16 at 2:40

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, $n=k$.

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.

• I might add that this could only be done by computer and the round off errors would make it impossible to actually achieve such precision. This is solely an academic exercise. – John Wayland Bales Nov 25 '16 at 2:49
• Checking the result on my calculator shows that one actually gets an error of $1.0000091\times10^{-4}$ so, is my calculator off? Probably, but if not we need $n=93969$. – John Wayland Bales Nov 25 '16 at 2:55
• The answer was 242. – Robert Miller Nov 25 '16 at 2:59
• @keving If you leave out the $200$ in the numerator, then $242$ would be the correct answer. So if your book or your teacher claims that the answer is $242$ then the numerator is wrong and should be $1$. – John Wayland Bales Nov 25 '16 at 6:02