5.4 Comparison tests  (Page 2/7)

Therefore, for all $k\ge 1,$

Since $a_1+a_2+\text{⋯}+a_{N-1}$ is a finite number, we conclude that the sequence $\left\{S_k\right\}$ is bounded above. Therefore, $\left\{S_k\right\}$ is an increasing sequence that is bounded above. By the Monotone Convergence Theorem, we conclude that $\left\{S_k\right\}$ converges, and therefore the series ${\displaystyle \sum _{n=1}^{\infty }{a}_n}$ converges.

□

To use the comparison test to determine the convergence or divergence of a series ${\displaystyle \sum _{n=1}^{\infty }{a}_n},$ it is necessary to find a suitable series with which to compare it. Since we know the convergence properties of geometric series and p -series, these series are often used. If there exists an integer $N$ such that for all $n\ge N,$ each term $a_n$ is less than each corresponding term of a known convergent series, then ${\displaystyle \sum _{n=1}^{\infty }{a}_n}$ converges. Similarly, if there exists an integer $N$ such that for all $n\ge N,$ each term $a_n$ is greater than each corresponding term of a known divergent series, then ${\displaystyle \sum _{n=1}^{\infty }{a}_n}$ diverges.

Limit comparison test

The comparison test works nicely if we can find a comparable series satisfying the hypothesis of the test. However, sometimes finding an appropriate series can be difficult. Consider the series

It is natural to compare this series with the convergent series

However, this series does not satisfy the hypothesis necessary to use the comparison test because

for all integers $n\ge 2.$ Although we could look for a different series with which to compare ${\displaystyle \sum }_{n=2}^{\infty }1\text{/}(n^2-1),$ instead we show how we can use the limit comparison test    to compare

Let us examine the idea behind the limit comparison test. Consider two series ${\displaystyle \sum }_{n=1}^{\infty }{a}_n$ and ${\displaystyle \sum }_{n=1}^{\infty }{b}_n.$ with positive terms $a_n\text{and}{b}_n$ and evaluate

If

then, for $n$ sufficiently large, $a_n\approx L{b}_n.$ Therefore, either both series converge or both series diverge. For the series ${\displaystyle \sum }_{n=2}^{\infty }1\text{/}(n^2-1)$ and ${\displaystyle \sum }_{n=2}^{\infty }1\text{/}{n}^2,$ we see that

Since ${\displaystyle \sum }_{n=2}^{\infty }1\text{/}{n}^2$ converges, we conclude that

converges.

The limit comparison test can be used in two other cases. Suppose

In this case, $\left\{a_n\text{/}{b}_n\right\}$ is a bounded sequence. As a result, there exists a constant $M$ such that $a_n\le M{b}_n.$ Therefore, if ${\displaystyle \sum }_{n=1}^{\infty }{b}_n$ converges, then ${\displaystyle \sum }_{n=1}^{\infty }{a}_n$ converges. On the other hand, suppose

In this case, $\left\{a_n\text{/}{b}_n\right\}$ is an unbounded sequence. Therefore, for every constant $M$ there exists an integer $N$ such that $a_n\ge M{b}_n$ for all $n\ge N.$ Therefore, if ${\displaystyle \sum }_{n=1}^{\infty }{b}_n$ diverges, then ${\displaystyle \sum }_{n=1}^{\infty }{a}_n$ diverges as well.