sum [n=1,inf.] [a_n] and sum [n=m,inf.] [a_n] or converge or diverge together

[youkito89]

Prove that series

\(\displaystyle \displaystyle \sum_{n=1 }^{+\infty }a_{n}\) and \(\displaystyle \sum_{n=m }^{+\infty }a_{n}\) or converge together or diverge together. When they converge, find \(\displaystyle \alpha\) to we have \(\displaystyle \sum_{n=1}^{+\infty }a_{n}=\alpha + \sum_{n=m}^{+\infty }a_{n}\)

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I will prove that \(\displaystyle \sum_{n=1 }^{+\infty }a_{n}\) converges \(\displaystyle \Leftrightarrow\) \(\displaystyle \sum_{n=m }^{+\infty }a_{n}\) converges.


We have the nth partial sums of seri \(\displaystyle \sum_{n=1 }^{+\infty }a_{n}\) is \(\displaystyle s_{n}=\sum_{k=1 }^{n }a_{k}\) and the nth partial sums of seri \(\displaystyle \sum_{n=m }^{+\infty }a_{n}\) is \(\displaystyle t_{n+m-1}=\sum_{k=m }^{n+m-1 }a_{k}\) (or the nth partial sums of seri \(\displaystyle \sum_{n=m }^{+\infty }a_{n}\) is\(\displaystyle t_{n}=\sum_{k=m }^{n}a_{k}\) ? I think I have problem with this)


We prove that the sequence \(\displaystyle (s_{n})\) converges \(\displaystyle \Leftrightarrow\) sequence \(\displaystyle (t_{n+m-1})\) converges


First, suppose we have the sequence \(\displaystyle (s_{n})\) converges and \(\displaystyle \lim_{n\rightarrow \infty }s_{n}=b\). We have \(\displaystyle \forall \epsilon >0,\exists n_{0}\in \mathbb{N}, \forall n\geq n_{0}:|s_{n}-b|<\epsilon\). Cause \(\displaystyle n+m-1\geq n \geq n_{0}\) so \(\displaystyle |s_{n+m-1}-b|< \epsilon \Rightarrow |t_{n+m-1}+ \sum_{k=1 }^{m-1 }a_{k}-b|< \epsilon\).


Therefore \(\displaystyle \forall \epsilon >0,\exists n_{0}\in \mathbb{N}, \forall n\geq n_{0}:|t_{n+m-1}-( b-\sum_{k=1 }^{m-1 }a_{k})|<\epsilon\). So \(\displaystyle (t_{n+m-1})\) converges and \(\displaystyle \lim_{n\rightarrow \infty }t_{n+m-1}=b-\sum_{k=1 }^{m-1 }a_{k} \)


Second, suppose we have the sequence \(\displaystyle (t_{n})\) converges and \(\displaystyle \lim_{n\rightarrow \infty }t_{n}=b\). I'm stuck at this.


Please tell me where I was wrong. Thank you very much.

 

[Deleted member 4993]

Your posted-image is too fuzzy. Cannot read the problem.

 

[youkito89]

Your posted-image is too fuzzy. Cannot read the problem.

Sorry for this picture. How can I edit my thread? Thank you.

Prove that series

\(\displaystyle \displaystyle \sum_{n=1 }^{+\infty }a_{n}\) and \(\displaystyle \sum_{n=m }^{+\infty }a_{n}\) or converge together or diverge together. When they converge, find \(\displaystyle \alpha\) to we have \(\displaystyle \sum_{n=1}^{+\infty }a_{n}=\alpha + \sum_{n=m}^{+\infty }a_{n}\)

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I will prove that \(\displaystyle \sum_{n=1 }^{+\infty }a_{n}\) converges \(\displaystyle \Leftrightarrow\) \(\displaystyle \sum_{n=m }^{+\infty }a_{n}\) converges.


We have the nth partial sums of seri \(\displaystyle \sum_{n=1 }^{+\infty }a_{n}\) is \(\displaystyle s_{n}=\sum_{k=1 }^{n }a_{k}\) and the nth partial sums of seri \(\displaystyle \sum_{n=m }^{+\infty }a_{n}\) is \(\displaystyle t_{n+m-1}=\sum_{k=m }^{n+m-1 }a_{k}\) (or the nth partial sums of seri \(\displaystyle \sum_{n=m }^{+\infty }a_{n}\) is\(\displaystyle t_{n}=\sum_{k=m }^{n}a_{k}\) ? I think I have problem with this)


We prove that the sequence \(\displaystyle (s_{n})\) converges \(\displaystyle \Leftrightarrow\) sequence \(\displaystyle (t_{n+m-1})\) converges


First, suppose we have the sequence \(\displaystyle (s_{n})\) converges and \(\displaystyle \lim_{n\rightarrow \infty }s_{n}=b\). We have \(\displaystyle \forall \epsilon >0,\exists n_{0}\in \mathbb{N}, \forall n\geq n_{0}:|s_{n}-b|<\epsilon\). Cause \(\displaystyle n+m-1\geq n \geq n_{0}\) so \(\displaystyle |s_{n+m-1}-b|< \epsilon \Rightarrow |t_{n+m-1}+ \sum_{k=1 }^{m-1 }a_{k}-b|< \epsilon\).


Therefore \(\displaystyle \forall \epsilon >0,\exists n_{0}\in \mathbb{N}, \forall n\geq n_{0}:|t_{n+m-1}-( b-\sum_{k=1 }^{m-1 }a_{k})|<\epsilon\). So \(\displaystyle (t_{n+m-1})\) converges and \(\displaystyle \lim_{n\rightarrow \infty }t_{n+m-1}=b-\sum_{k=1 }^{m-1 }a_{k} \)


Second, suppose we have the sequence \(\displaystyle (t_{n})\) converges and \(\displaystyle \lim_{n\rightarrow \infty }t_{n}=b\). I'm stuck at this.


Please tell me where I was wrong. Thank you very much.

 

[Ishuda]

Prove that series

\(\displaystyle \displaystyle \sum_{n=1 }^{+\infty }a_{n}\) and \(\displaystyle \sum_{n=m }^{+\infty }a_{n}\) or converge together or diverge together. When they converge, find \(\displaystyle \alpha\) to we have \(\displaystyle \sum_{n=1}^{+\infty }a_{n}=\alpha + \sum_{n=m}^{+\infty }a_{n}\)

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...

First, I believe you have the right idea but you have mixed up your indices a bit. Next, I think I would use the lemma (or whatever) which says something like "if the limit as n goes to infinity of A_n is A then the limit as n goes to infinity of A_n+B is A+B". That is, assume that \(\displaystyle \forall\, \epsilon\, >\, 0,\, \exists\, n_{0}\in \mathbb{N}\, \mathbb{|}\, \forall\, n\, \geq\, n_{0}\, |A_{n}-A|<\epsilon\), i.e.
A_n converges. Then for n>n0
\(\displaystyle -\epsilon\, \le\, A_n-A\, \le\, \epsilon\, \)
\(\displaystyle \implies\, -\epsilon\, \le\, (A_n+B)-(A+B)\, \le\, \epsilon\, \)
\(\displaystyle \implies\, |(A_n+B)-(A+B)|\, \lt\, \epsilon\)
That is A_n+B converges to A+B

Now let
B = \(\displaystyle \Sigma_{k=1}^{k=m-1} a_k\),
\(\displaystyle s_n\, =\, \Sigma_{k=1}^{k=n} a_k\),
\(\displaystyle t_n\, =\, \Sigma_{k=m}^{k=n} a_k\).
Then
t_n = s_n - B
so that if t_n converges, so does s_n. And
s_n = t_n + B
so that if s_n converges, so does t_n. That is t_n converges iff s_n converges.

 

[pka]

Prove that series
\(\displaystyle \displaystyle \sum_{n=1 }^{+\infty }a_{n}\) and \(\displaystyle \sum_{n=m }^{+\infty }a_{n}\) or converge together or diverge together. When they converge, find \(\displaystyle \alpha\) to we have \(\displaystyle \sum_{n=1}^{+\infty }a_{n}=\alpha + \sum_{n=m}^{+\infty }a_{n}\)

Frankly I find this tread confusing. I think this is a prefect example of the need for quantifiers.
For example, does the question mean that \(\displaystyle \large m\) fixed? In which case \(\displaystyle \exists m\).
Or does the question mean for every \(\displaystyle \large m\)? In which case \(\displaystyle \forall m\).

If it is the latter, then no one \(\displaystyle \large\bf{\alpha}\) will work.

We can define \(\displaystyle \alpha_m =\displaystyle\sum_{n=1}^{m}a_{n}\) then \(\displaystyle \displaystyle\sum_{n=1}^{\infty }a_{n}=\alpha_{m} + \sum_{n=m+1}^{\infty }a_{n}\).

 

[Ishuda]

Frankly I find this tread confusing. I think this is a prefect example of the need for quantifiers.
For example, does the question mean that \(\displaystyle \large m\) fixed? In which case \(\displaystyle \exists m\).
Or does the question mean for every \(\displaystyle \large m\)? In which case \(\displaystyle \forall m\).

If it is the latter, then no one \(\displaystyle \large\bf{\alpha}\) will work.

We can define \(\displaystyle \alpha_m =\displaystyle\sum_{n=1}^{m}a_{n}\) then \(\displaystyle \displaystyle\sum_{n=1}^{\infty }a_{n}=\alpha_{m} + \sum_{n=m+1}^{\infty }a_{n}\).

Sort of the difference between "uniform convergence" and "convergence". Thanks for bringing it up. I though of that when I gave the above answer but came to the conclusion the answer required an answer for a fixed m [still the \(\displaystyle \forall\)m type but for fixed m]. One reason is probably I would have had to think about it a bit more if the problem was of the "uniform convergence" type [still the \(\displaystyle \forall\)m type but not necessarily for fixed m]. Something I used to get hung up on sometimes and still do in my dotage.