Infinite Series .... Sohrab Exercise 2.3.10 (1) .... ....

[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 In order to formulate a rigorous proof to the proposition stated in Exercise 2.3.10 (1) ... ...

Exercise 2.3.10 (1) reads as follows:
View attachment 9062I am unsure of what would constitute a formal and rigorous proof to the proposition or statement of Exercise 2.3.10 (1) ...

Here is my attempt at a formal and rigorous proof ...Assume $$\sum_{ n = 1 }^{ \infty } x_n $$ is convergent ... then $$\sum_{ n = 1 }^{ \infty } x_n $$ is Cauchy ...Thus we have the following ...... for every $$\epsilon \gt 0 \ \exists \ N \text{ such that } m_0 \geq n_0 \geq N \ \Longrightarrow \ \vert s_{ m_0 } - s_{ n_0 } \vert \lt \epsilon$$ ... ...... that is ... for every $$\epsilon \gt 0 \ \exists \ N \text{ such that } m_0 \geq n_0 \geq N \ \Longrightarrow \ \left\vert \sum_{ k = n_0 + 1 }^{ m_0 } x_n \right \vert \lt \epsilon$$ ... ...If $$n_0 \geq m$$ then we are done ...If $$n_0 \lt m_0$$ ... then take $$n_0 = m$$ and we are done ...Now assume $$\sum_{ n = m }^{ \infty } x_n $$ is convergent ... ... then ... for every $$\epsilon \gt 0 \ \exists \ N \text{ such that } m_0 \geq n_0 \geq m \geq N \ \Longrightarrow \ \left\vert \sum_{ k = n_0 + 1 }^{ m_0 } x_n \right \vert \lt \epsilon$$ ... ...Clearly the above N ensures that for $$\sum_{ n = 1 }^{ \infty } x_n $$ ...

... $$\epsilon \gt 0 \ \exists \ N \text{ such that } m_0 \geq n_0 \geq N \ \Longrightarrow \ \left\vert \sum_{ k = n_0 + 1 }^{ m_0 } x_n \right \vert \lt \epsilon$$ ... ...

... that is $$\sum_{ n = 1 }^{ \infty }x_n $$ is convergent ...

Could someone please indicate whether the above proof is correct and acceptable ,,, indeed ... have I missed the point ...

Further could someone critique the above proof pointing out errors and deficiencies ...Help will be appreciated ...

Peter

 

[Olinguito]

$\displaystyle\sum_{n=1}^\infty x_n$ is convergent​

$\displaystyle\iff\ \lim_{k\to\infty}\sum_{n=1}^k x_n$ exists and is finite

$\displaystyle\iff\ \lim_{k\to\infty}\sum_{n=m}^k x_n$ $=$ $\displaystyle\lim_{k\to\infty}\left[\left(\sum_{n=1}^k x_n\right)-\sum_{n=1}^{m-1}x_n\right]$ $=$ $\displaystyle\left[\lim_{k\to\infty}\left(\sum_{n=1}^k x_n\right)\right]-\sum_{n=1}^{m-1}x_n$ exists and is finite

$\displaystyle\iff\ \sum_{n=m}^\infty x_n$ is convergent.

In other words, adding a finite number of terms to or subtracting a finite number of terms from a convergent series results in convergent series.

 

[Math Amateur]

$\displaystyle\sum_{n=1}^\infty x_n$ is convergent​

$\displaystyle\iff\ \lim_{k\to\infty}\sum_{n=1}^k x_n$ exists and is finite

$\displaystyle\iff\ \lim_{k\to\infty}\sum_{n=m}^k x_n$ $=$ $\displaystyle\lim_{k\to\infty}\left[\left(\sum_{n=1}^k x_n\right)-\sum_{n=1}^{m-1}x_n\right]$ $=$ $\displaystyle\left[\lim_{k\to\infty}\left(\sum_{n=1}^k x_n\right)\right]-\sum_{n=1}^{m-1}x_n$ exists and is finite

$\displaystyle\iff\ \sum_{n=m}^\infty x_n$ is convergent.

In other words, adding a finite number of terms to or subtracting a finite number of terms from a convergent series results in convergent series.

Thanks for the help Olinguito ... neat clear proof!

Peter