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

youkito89

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Prove that series

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

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


We have the nth partial sums of seri n=1+an\displaystyle \sum_{n=1 }^{+\infty }a_{n} is sn=k=1nak\displaystyle s_{n}=\sum_{k=1 }^{n }a_{k} and the nth partial sums of seri n=m+an\displaystyle \sum_{n=m }^{+\infty }a_{n} is tn+m1=k=mn+m1ak\displaystyle t_{n+m-1}=\sum_{k=m }^{n+m-1 }a_{k} (or the nth partial sums of seri n=m+an\displaystyle \sum_{n=m }^{+\infty }a_{n} istn=k=mnak\displaystyle t_{n}=\sum_{k=m }^{n}a_{k} ? I think I have problem with this)


We prove that the sequence (sn)\displaystyle (s_{n}) converges \displaystyle \Leftrightarrow sequence (tn+m1)\displaystyle (t_{n+m-1}) converges


First, suppose we have the sequence (sn)\displaystyle (s_{n}) converges and limnsn=b\displaystyle \lim_{n\rightarrow \infty }s_{n}=b. We have ϵ>0,n0N,nn0:snb<ϵ\displaystyle \forall \epsilon >0,\exists n_{0}\in \mathbb{N}, \forall n\geq n_{0}:|s_{n}-b|<\epsilon. Cause n+m1nn0\displaystyle n+m-1\geq n \geq n_{0} so sn+m1b<ϵtn+m1+k=1m1akb<ϵ\displaystyle |s_{n+m-1}-b|< \epsilon \Rightarrow |t_{n+m-1}+ \sum_{k=1 }^{m-1 }a_{k}-b|< \epsilon.


Therefore ϵ>0,n0N,nn0:tn+m1(bk=1m1ak)<ϵ\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 (tn+m1)\displaystyle (t_{n+m-1}) converges and limntn+m1=bk=1m1ak\displaystyle \lim_{n\rightarrow \infty }t_{n+m-1}=b-\sum_{k=1 }^{m-1 }a_{k}


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


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

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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

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

----------------------

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


We have the nth partial sums of seri n=1+an\displaystyle \sum_{n=1 }^{+\infty }a_{n} is sn=k=1nak\displaystyle s_{n}=\sum_{k=1 }^{n }a_{k} and the nth partial sums of seri n=m+an\displaystyle \sum_{n=m }^{+\infty }a_{n} is tn+m1=k=mn+m1ak\displaystyle t_{n+m-1}=\sum_{k=m }^{n+m-1 }a_{k} (or the nth partial sums of seri n=m+an\displaystyle \sum_{n=m }^{+\infty }a_{n} istn=k=mnak\displaystyle t_{n}=\sum_{k=m }^{n}a_{k} ? I think I have problem with this)


We prove that the sequence (sn)\displaystyle (s_{n}) converges \displaystyle \Leftrightarrow sequence (tn+m1)\displaystyle (t_{n+m-1}) converges


First, suppose we have the sequence (sn)\displaystyle (s_{n}) converges and limnsn=b\displaystyle \lim_{n\rightarrow \infty }s_{n}=b. We have ϵ>0,n0N,nn0:snb<ϵ\displaystyle \forall \epsilon >0,\exists n_{0}\in \mathbb{N}, \forall n\geq n_{0}:|s_{n}-b|<\epsilon. Cause n+m1nn0\displaystyle n+m-1\geq n \geq n_{0} so sn+m1b<ϵtn+m1+k=1m1akb<ϵ\displaystyle |s_{n+m-1}-b|< \epsilon \Rightarrow |t_{n+m-1}+ \sum_{k=1 }^{m-1 }a_{k}-b|< \epsilon.


Therefore ϵ>0,n0N,nn0:tn+m1(bk=1m1ak)<ϵ\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 (tn+m1)\displaystyle (t_{n+m-1}) converges and limntn+m1=bk=1m1ak\displaystyle \lim_{n\rightarrow \infty }t_{n+m-1}=b-\sum_{k=1 }^{m-1 }a_{k}


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


Please tell me where I was wrong. Thank you very much.
 
Last edited:
Prove that series

n=1+an\displaystyle \displaystyle \sum_{n=1 }^{+\infty }a_{n} and n=m+an\displaystyle \sum_{n=m }^{+\infty }a_{n} or converge together or diverge together. When they converge, find α\displaystyle \alpha to we have n=1+an=α+n=m+an\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 An is A then the limit as n goes to infinity of An+B is A+B". That is, assume that ϵ>0,n0Nnn0AnA<ϵ\displaystyle \forall\, \epsilon\, >\, 0,\, \exists\, n_{0}\in \mathbb{N}\, \mathbb{|}\, \forall\, n\, \geq\, n_{0}\, |A_{n}-A|<\epsilon, i.e.
An converges. Then for n>n0
ϵAnAϵ\displaystyle -\epsilon\, \le\, A_n-A\, \le\, \epsilon\,
     ϵ(An+B)(A+B)ϵ\displaystyle \implies\, -\epsilon\, \le\, (A_n+B)-(A+B)\, \le\, \epsilon\,
     (An+B)(A+B)<ϵ\displaystyle \implies\, |(A_n+B)-(A+B)|\, \lt\, \epsilon
That is An+B converges to A+B

Now let
B = Σk=1k=m1ak\displaystyle \Sigma_{k=1}^{k=m-1} a_k,
sn=Σk=1k=nak\displaystyle s_n\, =\, \Sigma_{k=1}^{k=n} a_k,
tn=Σk=mk=nak\displaystyle t_n\, =\, \Sigma_{k=m}^{k=n} a_k.
Then
tn = sn - B
so that if tn converges, so does sn. And
sn = tn + B
so that if sn converges, so does tn. That is tn converges iff sn converges.
 
Prove that series
n=1+an\displaystyle \displaystyle \sum_{n=1 }^{+\infty }a_{n} and n=m+an\displaystyle \sum_{n=m }^{+\infty }a_{n} or converge together or diverge together. When they converge, find α\displaystyle \alpha to we have n=1+an=α+n=m+an\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 m\displaystyle \large m fixed? In which case m\displaystyle \exists m.
Or does the question mean for every m\displaystyle \large m? In which case m\displaystyle \forall m.

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

We can define αm=n=1man\displaystyle \alpha_m =\displaystyle\sum_{n=1}^{m}a_{n} then n=1an=αm+n=m+1an\displaystyle \displaystyle\sum_{n=1}^{\infty }a_{n}=\alpha_{m} + \sum_{n=m+1}^{\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 m\displaystyle \large m fixed? In which case m\displaystyle \exists m.
Or does the question mean for every m\displaystyle \large m? In which case m\displaystyle \forall m.

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

We can define αm=n=1man\displaystyle \alpha_m =\displaystyle\sum_{n=1}^{m}a_{n} then n=1an=αm+n=m+1an\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 \forallm 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 \forallm type but not necessarily for fixed m]. Something I used to get hung up on sometimes and still do in my dotage.
 
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