Math Amateur
Gold Member
MHB
- 3,920
- 48
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
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