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The question is:
How many terms of the series $$ \frac{1}{(2)(3)(4)}-\frac{5}{(6)(7)(8)}+\frac{9}{(10)(11)(12)}-\frac{13}{(14)(15)(16)}+... $$ must be used to ensure that the error in the approximation of the limit of the sum does not exceed 0.00001?
The solution then states
The general term of the series is $$ \frac{(-1)^{n-1}(4n-3)}{(4n-2)(4n-1)4n} $$
1. What is the point behind the $(-1)^{n-1}$ part? Is this not always equal to -1 since:
$(-1)^{1-1}=-1$
$(-1)^{2-1}=-1$
$(-1)^{3-1}=-1$
$(-1)^{4-1}=-1$
$(-1)^{1000-1}=-1$?

Correction The error I made here was not trusting what I knew and then testing it on my calculator. The calculator however ran its calculation as $-(1^{n-1})$ instead of $(-1)^{n-1}$.

So that would then yeald the series $$ \frac{1}{(2)(3)(4)}-\frac{5}{(6)(7)(8)}\color{red}{+}\frac{9}{(10)(11)(12)}-\frac{13}{(14)(15)(16)}\color{red}{+}... $$

The solution then continues stating:
Let $u_n=\frac{(4n-3)}{(4n-2)(4n-1)4n}$. Since the series alternates, $u_{n+1}<u_n$, and $\lim_{n\rightarrow\infty}u_n=0$,....
2. How can $\lim_{n\rightarrow\infty}u_n=0$. Surely $\lim_{n\rightarrow\infty}u_n=1$, no?

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2  
$$(-1)^n=1\,\,\,\forall\,\,\text{even}\,\,n\in\Bbb Z $$ –  DonAntonio Nov 18 '12 at 12:12
    
You should know that an alternating series (one with alternation signs) in which the terms tend to zero is convergent, and the error in the partial sum is bounded by the absolute value of the first term omitted. If you don't know this you should find out how to prove it before you use it. –  Mark Bennet Nov 18 '12 at 13:35

2 Answers 2

up vote 3 down vote accepted

$$(1)\;\;\;\;\;\;\;\;\;\;(-1)^n=1\,\,\,\forall\,\,\text{even}\,\,n\in\Bbb Z $$

$$(2)\;\;\;\;\;\;\;\;\;\;\;\frac{(-1)^{n-1}(4n-3)}{(4n-2)(4n-1)4n}\xrightarrow [n\to\infty]{} 0$$

since the denominator's degree in $\,n\,$ is higher than the numerator's

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1  
What happend to the -1 in $(-1)^{n\color{red}{-1}}$? –  Gineer Nov 18 '12 at 12:23
2  
What's the diffence?! As $\,n\,$ runs over the naturals, or even the integers, so does $\,n-1\,$. Fo example, $\,n-1\,$ is even for $\,n=1,3,5,7,9,...\,$ –  DonAntonio Nov 18 '12 at 12:26
    
@Gineer how did you put the red color? –  doniyor Nov 18 '12 at 15:29

$(-1)^{n-1}=1 \ \forall$ ODD $n$.

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