I would like to receive some help about the next problem:

The problem

I am trying to understand the proof of the following theorem about the two real series with positive members:

"Let it exist $\lim_{n \to +\infty} \frac{a_n}{b_n} = K$, $0 \le K \le +\infty$, where $a_n$ and $b_n$ are members of the real series with positive numbers $\sum_{n = 0}^{+\infty} a_n$ and $\sum_{n = 0}^{+\infty} b_n$, respectively.

If $K < +\infty$, then, from the convergence of the series $\sum_{n = 0}^{+\infty} b_n$ follows convergence of the series $\sum_{n = 0}^{+\infty} a_n$.

If $K > 0$, then, from the divergence of the series $\sum_{n = 0}^{+\infty} b_n$ follows divergence of the series $\sum_{n = 0}^{+\infty} a_n$."

The proof starts like this:

"Let $K < +\infty$ and let series $\sum_{n = 0}^{+\infty} b_n$ converge. Now, we have that for arbitrarily $\varepsilon > 0$ there exists $n_0 \in \mathbb{N}$, such that $\left| \frac{a_n}{b_n} - K \right| < \varepsilon$, for $n > n_0$ (I understand this.). From here we get the next inequality: $$(1) \quad 0 \le a_n < (K + \varepsilon)b_n, \qquad for \quad n > n_0."$$

Unfortunately, i don't understand how did we got the inequality (1). I tried to brake starting inequality in cases and i ended up confused:

1) $\frac{a_n}{b_n} - K < 0$:

$$\left| \frac{a_n}{b_n} - K \right| < \varepsilon \iff -\left( \frac{a_n}{b_n} - K \right) < \varepsilon \iff$$

$$\iff \begin{cases} (K - \varepsilon)b_n < a_n, \quad b_n > 0, \\ (K - \varepsilon)b_n > a_n, \quad b_n < 0. \end{cases}$$

My question:

I didn't get the sum $K + \varepsilon$ and i currently don't have an idea how to get what is needed to continue the proof.

Presuming that inequality (1) is correct, than the rest of the proof is understandable to me, but i don't have a full connection between two parts.

Please, could you help me with some hint or advice on how to get the inequality (1)?

Thank you, for your help and your time!

  • 1
    $\begingroup$ First, we are told that $a_n$ and $b_n$ are positive, so you don't have to worry about the cases $a_n<0, b_n<0.$ The first inequality, $0\le a_n$ comes from the hypothesis; you don't have to worry about that either. Now try it again. $\endgroup$ – saulspatz Nov 16 '18 at 22:58
  • $\begingroup$ @saulspatz: I know that definition of the real series with positive members says that such series could contain finite number of negative members (up until $n_0$-member). I know that convergence doesn't change if we remove some finite number of members of the series. So, in this case i have the right to don't take those eventual negative members into consideration? $\endgroup$ – MathsLearner Nov 16 '18 at 23:10

Edit $$ \left| \frac{a_n}{b_n} - K \right| < \varepsilon\iff \left| \frac{a_n}{b_n} - K \right| +K < K+\varepsilon\label{A}\tag{A} $$ Then by the triangle inequality we have $$ \left| \frac{a_n}{b_n} - K \right| +K \ge\left| \frac{a_n}{b_n} - K +K \right| = \left| \frac{a_n}{b_n}\right|\label{B}\tag{B} $$ and using \eqref{A} and \eqref{B} we obtain $$ \left| \frac{a_n}{b_n}\right|<K+\varepsilon\iff 0 \le |a_n| < (K + \varepsilon)|b_n|, \quad \text{ for }\quad n > n_0\label{C}\tag{C} $$ Now since the coefficients of real series with positive numbers, $a_n,b_n\ge 0$ except for a finite number of them, say $$ \begin{split} a_n&\ge 0\quad\text{ for all }n>n_a\\ b_n&\ge 0\quad\text{ for all }n>n_b \end{split}\label{D}\tag{D} $$ Inequality \eqref{A} implies that $a_n/b_n>0$ for all $n>n_0$ since $K>0$ and by a standard lemma in mathematical analysis $$ \lim_{n\to\infty} \left| \frac{a_n}{b_n} - K \right|=0\iff\lim_{n\to\infty} \left( \frac{a_n}{b_n} - K \right) $$ This implies that $a_n$ and $b_n$ have the same sign for all $n>n_0$ thus, by defining $$ n_+=\max\{n_0, n_a, n_b\}, $$ we have that, for all $n>n_+$, the coefficients $a_n$ and $b_n$ are respectively non negative and positive real numbers so we can remove the absolute values from \eqref{C} and write $$ 0 \le a_n < (K + \varepsilon)b_n, \quad \text{ for }\quad n > n_+ $$

  • $\begingroup$ Thank you for your answer! So, in this case there is no need to take into consideration eventual negative members of the series, because, by definition, there might be a finite number of them in the positive series? $\endgroup$ – MathsLearner Nov 16 '18 at 23:14
  • $\begingroup$ No, there just aren't any at all: it's given in the statement of the theorem. $\endgroup$ – user3482749 Nov 16 '18 at 23:22
  • $\begingroup$ Exactly: from some index $n_0$ on, the have that the ratio $a_n/b_n$ is positive and the positivity of one of the coefficients implies that even the other is positive. $\endgroup$ – Daniele Tampieri Nov 16 '18 at 23:22
  • $\begingroup$ I am sorry, but i have to ask again. How do we, exactly, remove all possible negative (for $b_n$, because it can't be $0$) and non-positive (for $a_n$) members from consideration during the proof of this theorem? With what words would i explain that to someone? By the definition, $\sum_{n = 0}^{+\infty} a_n$ and $\sum_{n = 0}^{+\infty} b_n$, could contain finite number of negative members. Maybe they don't, but they could. $\endgroup$ – MathsLearner Nov 16 '18 at 23:41
  • 1
    $\begingroup$ @OgnjenMojovic: I have further modified my answer in order to precise all steps. I hope now it is sufficiently clear. $\endgroup$ – Daniele Tampieri Nov 17 '18 at 0:32

(1) is just a rearrangement of the inequality above:

$$\left|\frac{a_n}{b_n} - K\right| < \varepsilon$$ is equivalent to $$-\varepsilon < \frac{a_n}{b_n} - K < \varepsilon,$$ and (since $b_n$ is positive) multiplying everything in sight by $b_n$ gives $$-\varepsilon b_n < a_n - Kb_n < \varepsilon,$$ and adding $Kb_n$ to everything gives $$b_n(K - \varepsilon) < a_n < b_n(K + \varepsilon),$$

but we already know that $a_n$ is positive, so we have $$0 \leq a_n < b_n(K + \varepsilon).$$

  • $\begingroup$ Thank you for your answer! Please, could you tell me if there is a typo, because i don't understand why did you switch $K$ with $1$? $\endgroup$ – MathsLearner Nov 16 '18 at 23:19
  • 1
    $\begingroup$ Yes, I failed at typing. Will fix. $\endgroup$ – user3482749 Nov 16 '18 at 23:21

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