First Comparison Test for Series .... Sohrab Theorem 2.3.9 ....

[Math Amateur]

I am reading Houshang H. Sohrab's book: "Basic Real Analysis" (Second Edition).

I am focused on Chapter 2: Sequences and Series of Real Numbers ... ...

I need help with the proof of Theorem 2.3.9 (a) ...

Theorem 2.3.9 reads as follows:
View attachment 9065
Now, we can prove Theorem 2.3.9 (a) using the Cauchy Criterion for Series ... as follows:Since \(\displaystyle 0 \leq x_n \leq y_n \ \forall \ n \in \mathbb{N}\) ... we have that ...\(\displaystyle \left\vert \sum_{ k = n + 1 }^m x_n \right\vert \leq \left\vert \sum_{ k = n + 1 }^m y_n \right\vert\) ... ... ... ... ... (1)But ... since \(\displaystyle \sum_{ n = 1 }^{ \infty } y_n \) is convergent, it satisfies the Cauchy Criterion for Series ...... further ... (1) implies that \(\displaystyle \sum_{ n = 1 }^{ \infty } x_n \) also satisfies the Cauchy Criterion for Series ... ... and so \(\displaystyle \sum_{ n = 1 }^{ \infty } x_n \) is convergent ...

However ... Sohrab states that Theorem 2.3.9 (a) is an immediate consequence of Theorem 2.3.6 (see below) ... ... this means we need to establish bounds for \(\displaystyle \sum_{ n = 1 }^{ \infty } x_n \) ...... can someone please demonstrate how to do this ...
Note: If \(\displaystyle \lim_{n \to \infty } t_n = \lim_{n \to \infty } \sum_{ k = 1 }^n y_k = L\) ...... does \(\displaystyle L\) form an upper bound on \(\displaystyle \sum_{ n = 1 }^{ \infty } y_n \) ...... and hence also form an upper bound on \(\displaystyle \sum_{ n = 1 }^{ \infty } x_n\) ...... but ... how do we prove this ... ?

Help will be appreciated ...

Peter
=======================================================================================The post above refers to Theorem 2.3.6 ... so I am providing text of the same ... as follows:View attachment 9066
Hope that helps ...

Peter

 

[Euge]

Hi Peter,

First, let's talk about Theorem 2.3.6. Recall that an increasing sequence of real numbers converges if and only if it is bounded. Given a series $\sum x_n$ of nonnegative terms, its sequence of partial sums $s_N = \sum_1^N x_n$ is increasing. Since $s_N$ converges if and only if it is bounded and the sum of a series is defined as the limit the sequence of partial sums, it follows that $\sum x_n$ converges if and only if it is bounded.

Now, let's revisit the proof of Theorem 2.3.9. If $\sum y_n$ converges, then $t_n$ is bounded, say, by $T$. As $0 \le s_n \le t_n$ for all $n$, then $0 \le s_n \le T$ for all $n$. Thus $s_n$ is bounded, which implies $s_n$ converges. In other words, $\sum x_n$ converges. This proves $(a)$. Statement $(b)$ is equivalent to statement $(a)$.

 

[Math Amateur]

Hi Peter,

First, let's talk about Theorem 2.3.6. Recall that an increasing sequence of real numbers converges if and only if it is bounded. Given a series $\sum x_n$ of nonnegative terms, its sequence of partial sums $s_N = \sum_1^N x_n$ is increasing. Since $s_N$ converges if and only if it is bounded and the sum of a series is defined as the limit the sequence of partial sums, it follows that $\sum x_n$ converges if and only if it is bounded.

Now, let's revisit the proof of Theorem 2.3.9. If $\sum y_n$ converges, then $t_n$ is bounded, say, by $T$. As $0 \le s_n \le t_n$ for all $n$, then $0 \le s_n \le T$ for all $n$. Thus $s_n$ is bounded, which implies $s_n$ converges. In other words, $\sum x_n$ converges. This proves $(a)$. Statement $(b)$ is equivalent to statement $(a)$.

Thanks for the explanation, Euge ...

Most helpful!

Peter