20. Convergence of Positive Series

d.1. The Limit Comparison Test

a. Proof

Cases (2) and (3) are called the extreme cases, and arise very rarely.
If \(L=0\) and \(\displaystyle \sum_{n=n_o}^\infty b_n\) is divergent, or \(L=\infty\) and \(\displaystyle \sum_{n=n_o}^\infty b_n\) is convergent, the Limit Comparison Test FAILS.

We will only prove Case (1) where \(0 \lt L \lt \infty\). The statement \(\displaystyle \lim_{n\to\infty}\dfrac{a_n}{b_n}=L\) means:

For all   \(\varepsilon \gt 0\),   there is a positive integer,   \(N\),   such that

if   \(n \gt N\)   then   \(L-\varepsilon \lt \dfrac{a_n}{b_n} \lt L+\varepsilon\).

We pick \(\varepsilon=\dfrac{L}{2}\). Then there is an \(N\) such that if \(n \ge N\) then: \[ L-\dfrac{L}{2} \lt \dfrac{a_n}{b_n} \gt L+\dfrac{L}{2} \] Consequently, \[ \dfrac{L}{2}b_n \lt a_n \lt \dfrac{3L}{2}b_n \] Now we apply the Simple Comparison Test to the tail of the series starting at \(n=N\):

If \(\displaystyle \sum_{n=n_o}^\infty b_n\) is convergent, then \(\displaystyle \dfrac{3L}{2}\sum_{n=N}^\infty b_n\) is convergent and hence \(\displaystyle \sum_{n=N}^\infty a_n\) is convergent and so \(\displaystyle \sum_{n=n_o}^\infty a_n\) is convergent.

If \(\displaystyle \sum_{n=n_o}^\infty b_n\) is divergent, then \(\displaystyle \dfrac{L}{2}\sum_{n=N}^\infty b_n\) is divergent and hence \(\displaystyle \sum_{n=N}^\infty a_n\) is divergent and so \(\displaystyle \sum_{n=n_o}^\infty a_n\) is divergent.

If you understood this proof, try proving Cases (2) and (3). For Case (2), you should pick \(\varepsilon=1\). Case (3) can be proved using a proof by contradiction and the fact that if \(\displaystyle \lim_{n\to\infty}\dfrac{a_n}{b_n}=\infty\) then \(\displaystyle \lim_{n\to\infty}\dfrac{b_n}{a_n}=0\).