# Approximate the value of the series to within an error of at most 10^{-4}. \sum_{n=1}^{\infty...

## Question:

Approximate the value of the series to within an error of at most {eq}10^{-4}{/eq}.

{eq}\sum_{n=1}^{\infty }\frac{(-1)^{n+1}}{(n+80)(n+75)}{/eq}

According to the next equation:

{eq}\left | S_{N} - S \right | \leq a_{N+1}{/eq} what is the smallest value of {eq}N{/eq} that approximates {eq}S{/eq} to within an error of at most {eq}10^{-4}{/eq}?

## Remainder of series:

The remainder in a series refers to the error of the partial sum from the true sum of the series. The remainder is used to solve for the nth term minimum for a desired error from the true sum.

## Answer and Explanation:

Given the the series

$$\sum _{n=1}^{\infty }\frac{\left(-1\right)^{n+1}}{\left(n+80\right)\left(n+75\right)} $$

Then we need to solve for the {eq}|S_N - S| {/eq} such that it is within {eq}10^{-4} {/eq}.

$$\begin{align} 10^{-4}&=\frac{\left(-1\right)^{n+1}}{\left(n+80\right)\left(n+75\right)} \\ 0.0001&=\frac{\left(-1\right)^{n+1}}{\left(n+80\right)\left(n+75\right)} \\ 0.0001&=\frac{1}{\left(n+80\right)\left(n+75\right)} \\ n^2+155n+6000&=\frac{1}{0.0001} \\ n^2+155n+6000&=10000 \\ n^2+155n-4000&=0 \\ \end{align} $$

Using the quadratic formula

$$\begin{align} n&=\frac{-b\pm \sqrt{b^2-4ac}}{2a} \\ n&=\frac{-155+\sqrt{155^2-4\left(1\right)\left(-4000\right)}}{2\left(1\right)} \\ n&=22.53124 \\ n&=\frac{-155-\sqrt{155^2-4\left(1\right)\left(-4000\right)}}{2\left(1\right)} \\ n&=-177.53124 \end{align} $$

Round up

$$n=23 $$

Then, we know that

$$N = n-1\\ N = 23 - 1 $$

$$\boxed{N=22} $$

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from GRE Math: Study Guide & Test Prep

Chapter 12 / Lesson 4