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

In summary, Peter demonstrates that Theorem 2.3.9 is an immediate consequence of Theorem 2.3.6. Statements $(b)$ and $(c)$ are equivalent to statement $(a)$.
  • #1
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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
 

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  • #2
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 $(a)$.
 
  • #3
Euge said:
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
 

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

What is the First Comparison Test for Series?

The First Comparison Test for Series, also known as Sohrab Theorem 2.3.9, is a mathematical tool used to determine whether a series converges or diverges. It compares the given series to another series whose convergence or divergence is already known.

How is the First Comparison Test for Series used?

The First Comparison Test for Series is used by comparing the given series to another series with known convergence or divergence. If the known series converges and the given series is less than or equal to it, then the given series also converges. If the known series diverges and the given series is greater than or equal to it, then the given series also diverges.

What is the purpose of the First Comparison Test for Series?

The purpose of the First Comparison Test for Series is to determine the convergence or divergence of a given series. It is a useful tool in mathematical analysis and is often used to simplify complex series and make them easier to evaluate.

What are the limitations of the First Comparison Test for Series?

The First Comparison Test for Series can only be used for series with positive terms. Additionally, it can only determine whether a series converges or diverges, but it cannot determine the exact value of the sum of a convergent series.

Can the First Comparison Test for Series be used for all types of series?

No, the First Comparison Test for Series can only be used for series with positive terms. It cannot be used for series with alternating terms, such as alternating series or series with negative terms. In these cases, other convergence tests such as the Alternating Series Test or the Ratio Test should be used.

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