# 1. Determine N so that sum_{n=1}^{N} (- 1)^n {1/{2 n + 1}} differs from the sum of the series...

## Question:

1. Determine {eq}N {/eq} so that {eq}\sum_{n=1}^{N}(-1)^n\frac{1}{2n+1} {/eq} differs from the sum of the series {eq}\sum_{n=1}^{\infty}(-1)^n\frac{1}{2n+1} {/eq} by less than {eq}\frac{1}{1000} {/eq}.

2. Find the sum of the series {eq}\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n!} {/eq} correct to {eq}4 {/eq} decimals.

## Alternating series remainder:

The alternating series remainder theorem states that if an convergent alternating series is evaluated at some values of {eq}n {/eq}, then the absolute value of its remainder shall be less than the rule of series evaluated at {eq}n + 1 {/eq}.

## Answer and Explanation:

**1.**

Since we are asked to find for the {eq}N {/eq}th term such that the remainder is less than {eq}\frac{1}{1000} {/eq}, we know that the series converges. Using the theorem for alternating series remainder which

$$|R_n| \leq a_{n+1} $$

We want our {eq}|R_n| = \frac{1}{1000} {/eq}

$$\begin{align} \frac{1}{1000} &\leq \left(-1\right)^n\frac{1}{2n+1} \\ \frac{1}{1000}&=\left(-1\right)^n\frac{1}{2n+1} \\ \frac{1}{1000}&=\frac{1}{2n+1} \\ 2n+1&=1000 \\ n&=\frac{999}{2} \\ n &= 499.5 \end{align} $$

Rounding up

$$n = 500 $$

Now, remember that the theorem is comparing to the {eq}a_{n+1} {/eq} term. Hence, {eq}N = 500 - 1 = 499 {/eq}

$$\boxed{N = 499} $$

**2.**

Using the same theorem

$$|R_n| \leq a_{n+1} $$

We want our {eq}|R_n| = 0.00005 {/eq}

$$\begin{align} 0.00005 &\leq \frac{\left(-1\right)^{n-1}}{n!} \\ 0.00005&=\frac{\left(-1\right)^{n-1}}{n!} \\ 0.00005&=\frac{1}{n!} \\ n!&=\frac{1}{0.00005} \\ n!&=20000 \\ \end{align} $$

Since {eq}7! = 5040 {/eq} and {eq}8! = 40320 {/eq}. Then we will choose {eq}n = 8 {/eq}. Hence, {eq}N = 8 - 1 = 7 {/eq}.

$$\boxed{N = 7} $$

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from GRE Math: Study Guide & Test Prep

Chapter 12 / Lesson 4