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I'm studying on my own using Bonar and Khoury's Real Infinite Series. I understand the proof of the "regular" Limit Comparison Test( a link to google books, page 23 ) but the book doesn't provide a proof for the strengthened version, which is on the next page on the link, stated as

Let $\sum_{n=0}^\infty a_n$ and $\sum_{n=0}^\infty b_n $ be two strictly positive series, and suppose that $$\liminf_{n\to\infty} \frac{a_n}{b_n}=L_1\ \ ,\ \ \limsup_{n\to\infty} \frac{a_n}{b_n}=L_2 $$ then

1.If $L_2 \lt \infty$ and $\sum_{n=0}^\infty b_n $ converges, then $\sum_{n=0}^\infty a_n$ also converges

If $L_1 \gt 0$ and $\sum_{n=0}^\infty b_n $ diverges, then $\sum_{n=0}^\infty a_n$ also diverges.

Request: Can you provide a proof for this theorem? And can you please provide it in a way that stresses on understanding of the limit superior and inferior concept? It what I am having trouble getting my head around. Thanks in advance.

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  • $\begingroup$ $\limsup \frac{a_n}{b_n} < +\infty$ means (under the given conditions) that there is a $C \in (0,+\infty)$ with $a_n \leqslant C\cdot b_n$ for all $n$. Can you continue? $\endgroup$ – Daniel Fischer Apr 27 '16 at 18:38
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To say that $\displaystyle \limsup_{n \to \infty} \frac{a_n}{b_n} = L_2$ means: first, for any positive $\epsilon > 0$ there exists $N$ with the property that $n \ge N$ implies $\dfrac{a_n}{b_n} < L_2 + \epsilon$, and second, $L_2$ is the least number with this property.

This is more than is required. It suffices to know that for some $N$, $\dfrac{a_n}{b_n} < 2L_2$ for all $n \ge N$. Then $$n \ge N \implies a_n < 2L_2 b_n,$$ and since $\displaystyle \sum_{n=N}^\infty 2L_2 b_n$ converges, so does $\displaystyle \sum_{n=N}^\infty a_n$ by the usual comparison test. Adding the first $N-1$ terms does not affect convergence.

You can do the other problem using the comparison test for divergence and use $L_1/2$ as the lower bound of the ratio.

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