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### Section 4-8 : Alternating Series Test

5. Determine if the following series converges or diverges.

$\sum\limits_{n = 4}^\infty {\frac{{{{\left( { - 1} \right)}^{n + 2}}\left( {1 - n} \right)}}{{3n - {n^2}}}}$

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First, this is (hopefully) clearly an alternating series with,

${b_n} = \frac{{1 - n}}{{3n - {n^2}}}$

and $${b_n}$$ are positive for $$n \ge 4$$ and so we know that we can use the Alternating Series Test on this series.

It is very important to always check the conditions for a particular series test prior to actually using the test. One of the biggest mistakes that many students make with the series test is using a test on a series that don’t meet the conditions for the test and getting the wrong answer because of that!

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Let’s first take a look at the limit,

$\mathop {\lim }\limits_{n \to \infty } {b_n} = \mathop {\lim }\limits_{n \to \infty } \frac{{1 - n}}{{3n - {n^2}}} = 0$

So, the limit is zero and so the first condition is met.

Show Step 3

Now let’s take care of the decreasing check. In this case increasing $$n$$ will increase both the numerator and denominator and so we can’t just say that clearly the terms are decreasing as we did in the first few problems.

We will have no choice but to do a little Calculus I work for this problem. Here is the function and derivative for that work.

$f\left( x \right) = \frac{{1 - x}}{{3x - {x^2}}}\hspace{0.5in}f'\left( x \right) = \frac{{ - {x^2} + 2x - 3}}{{{{\left( {3x - {x^2}} \right)}^2}}}$

The numerator of the derivative is never zero for any real number (we’ll leave that to you to verify) and since it is clearly negative at $$x = 0$$ we know that the function will always be decreasing for $$x \ge 4$$.

Therefore the $${b_n}$$ form a decreasing sequence.

Show Step 4

So, both of the conditions in the Alternating Series Test are met and so the series is convergent.