If a series converges, then the limit of the terms in the sequence must approach 0; if not, the series will diverge. The author also notes that adding or subtracting any finite number of terms from a series does not change its convergence or divergence property.

To better answer the query about using algebra to break down and evaluate series, let’s consider an example:

Example 8.2.6 (revised): Evaluate the given series.
1. n=1(1)n+1(n^2-n)n^3/2
Solution:
We can use algebra to break down this series and evaluate its convergence or divergence. First, we’ll simplify the expression inside the parentheses:
(n^2-n)n^3/2 = n^5/2 (1/2)n^4
Next, let’s apply the comparison test for convergence of series to see if this series converges or diverges. We can compare it with a known convergent series, such as n=1(1/n^3). If our series is less than that series in absolute value (i.e., |(1)n+1(n^5/2 (1/2)n^4)| < |1/n^3|), then it converges by the comparison test. To do this, we'll take the absolute values of both sides and simplify: |(1)n+1(n^5/2 (1/2)n^4)| = |(1)^n*(n-1)(n^3(n-2))/(2*n)| To make this comparison easier, let's factor out the absolute value signs and simplify: |(1)n+1(n^5/2 (1/2)n^4)| = |(n-1)(n-2)/(2*n)|*(n^3)*|(1)^n| Now, let's compare this to the known convergent series: |(n-1)(n-2)/(2*n)| < 1/n^2 (for n > 2)
If we can show that |(n-1)(n-2)/(2*n)| is less than or equal to 1/n^2 for all values of n greater than 2, then our series will converge by the comparison test. To do this, let’s divide both sides by (n-2) and simplify:
|(n-1)/(2*n)| < |1/(n-2)*1/n^2| To make this easier to compare with 1/n^2, we can take the limit as n approaches infinity of both sides. This gives us: lim_{n}|(n-1)/(2*n)| = lim_{n} |1/(2*(n/n))| * lim_{n} |1/n^2| = 0 * 0 = 0 Since the limit is less than or equal to zero, we can say that for all values of n greater than 2: |(n-1)/(2*n)| < |1/(n-2)*1/n^2| This means that our series converges by the comparison test. Therefore, this particular series will also converge.