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January 2nd, 2016, 08:54 AM  #1 
Newbie Joined: Jan 2016 From: Viet Nam Posts: 3 Thanks: 0  $\sum_{n=1 }^{+âˆž }a_{n}$ and $\sum_{n=m }^{+âˆž }a_{n}$ converge or diverge together
Prove that series $$\sum_{n=1 }^{+\infty }a_{n}$$ and $$\sum_{n=m }^{+\infty }a_{n}$$ or converge together or diverge together. When they converge, find $\alpha$ to we have $$\sum_{n=1}^{+\infty }a_{n}=\alpha + \sum_{n=m}^{+\infty }a_{n}$$ *** I will prove that $\sum_{n=1 }^{+\infty }a_{n}$ converges $\Leftrightarrow$ $\sum_{n=m }^{+\infty }a_{n}$ converges. We have the nth partial sums of seri $\sum_{n=1 }^{+\infty }a_{n}$ is $s_{n}=\sum_{k=1 }^{n }a_{k}$ and the nth partial sums of seri $\sum_{n=m }^{+\infty }a_{n}$ is $t_{n+m1}=\sum_{k=m }^{n+m1 }a_{k}$ ***(or the nth partial sums of seri $\sum_{n=m }^{+\infty }a_{n}$ is$t_{n}=\sum_{k=m }^{n}a_{k}$ ? I think I have problem with this)*** We prove that the sequence $(s_{n})$ converges $\Leftrightarrow$ sequence $(t_{n+m1})$ converges First, suppose we have the sequence $(s_{n})$ converges and $\lim_{n\rightarrow \infty }s_{n}=b$. We have $\forall \epsilon >0,\exists n_{0}\in \mathbb{N}, \forall n\geq n_{0}: s_{n}b<\epsilon$. Cause $n+m1\geq n \geq n_{0}$ so $s_{n+m1}b< \epsilon \Rightarrow t_{n+m1}+ \sum_{k=1 }^{m1 }a_{k}b< \epsilon$. Therefore $\forall \epsilon >0,\exists n_{0}\in \mathbb{N}, \forall n\geq n_{0} : t_{n+m1}( b\sum_{k=1 }^{m1 }a_{k})<\epsilon$. So $(t_{n+m1})$ converges and $\lim_{n\rightarrow \infty }t_{n+m1}=b\sum_{k=1 }^{m1 }a_{k} $ Second, suppose we have the sequence $(t_{n})$ converges and $\lim_{n\rightarrow \infty }t_{n}=b$. I'm stuck at this. Please tell me where I was wrong. Thank you very much. 
January 2nd, 2016, 01:38 PM  #2 
Math Team Joined: Nov 2014 From: Australia Posts: 689 Thanks: 244 
Your work for the first part looks fine, although a different and easier way to start would be $$\sum_{n = 1}^\infty a_n = \sum_{n = m}^\infty a_n + \sum_{n = 1}^{m  1}a_n.$$ The second sum on the RHS is finite, so there are no convergence issues. Now just look at the cases where each series converges and diverges. As for your concerns in brackets, I think defining $\displaystyle t_n = \sum_{k = m}^n a_k$ would be slightly easier. Then you could go immediately from $$s_n  b < \epsilon$$ to $$t_n + \sum_{k = 1}^{m  1}a_k  b < \epsilon$$ 
January 2nd, 2016, 07:21 PM  #3  
Newbie Joined: Jan 2016 From: Viet Nam Posts: 3 Thanks: 0  Quote:
Because $\sum_{n=1 }^{m1 }a_{n}$ is a finite sum (constant) so $\sum_{n=1 }^{\infty}a_{n}$ converges $\Leftrightarrow$ $\sum_{n=m }^{\infty}a_{n}$ converges.  By the way, I have a question about the nth partial sum, must it have n terms? Like the nth partial sum of $\sum_{n = 1}^\infty a_n$ is $\sum_{k = 1}^n a_k$ and it has n terms. Therefore the nth partial sum of $\sum_{n = m}^\infty a_n$ is $\sum_{k = m}^{n+m1} a_k$ and it has n terms. As you said that the nth partial sum of $\sum_{n = m}^\infty a_n$ is $\sum_{k = m}^{n} a_k$, so which way is right? Or both right? I'm so confused about this.  
January 3rd, 2016, 01:31 AM  #4  
Math Team Joined: Nov 2014 From: Australia Posts: 689 Thanks: 244 
Don't quote me on this, but I believe the $n$th partial sum of a sequence $\{a_n\}$ is defined as $$s_n = \sum_{k = 1}^n a_n.$$ That is, the sum of the first $n$ terms of $\{a_n\}$. I don't think $\displaystyle\sum_{k = m}^{n + m  1} a_n$ counts as a partial sum at all. Quote:
 
January 3rd, 2016, 02:11 AM  #5 
Newbie Joined: Jan 2016 From: Viet Nam Posts: 3 Thanks: 0 
So the nth partial sum of $\sum_{n = m}^\infty a_n$ is $\sum_{k = m}^{n} a_k$ . If this, I can complete this exercise. Thank you.

January 3rd, 2016, 12:58 PM  #6 
Math Team Joined: Nov 2014 From: Australia Posts: 689 Thanks: 244 
If I'm correct, there's no such thing as the $n$th partial sum of $\displaystyle\sum_{n = m}^\infty a_n$. That doesn't make much difference to your proof though. You can still let $\displaystyle t_n = \sum_{k = m}^n a_n$ and it works fine. Even if it isn't an $n$th partial sum. 

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$sumn1, $sumnm, an$, âˆž, calculus, converge, converge or diverge, diverge, infty, partial sums, seri 
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