My Math Forum $\sum_{n=1 }^{+âˆž }a_{n}$ and $\sum_{n=m }^{+âˆž }a_{n}$ converge or diverge together

 Calculus Calculus Math Forum

 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 $t_{n+m-1}=\sum_{k=m }^{n+m-1 }a_{k}$ ***(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 $(t_{n+m-1})$ 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 $n+m-1\geq n \geq n_{0}$ so $|s_{n+m-1}-b|< \epsilon \Rightarrow |t_{n+m-1}+ \sum_{k=1 }^{m-1 }a_{k}-b|< \epsilon$. Therefore $\forall \epsilon >0,\exists n_{0}\in \mathbb{N}, \forall n\geq n_{0} : |t_{n+m-1}-( b-\sum_{k=1 }^{m-1 }a_{k})|<\epsilon$. So $(t_{n+m-1})$ converges and $\lim_{n\rightarrow \infty }t_{n+m-1}=b-\sum_{k=1 }^{m-1 }a_{k}$ 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 $$|t_n + \sum_{k = 1}^{m - 1}a_k - b| < \epsilon$$ Thanks from youkito89
January 2nd, 2016, 07:21 PM   #3
Newbie

Joined: Jan 2016
From: Viet Nam

Posts: 3
Thanks: 0

Quote:
 Originally Posted by Azzajazz 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 $$|t_n + \sum_{k = 1}^{m - 1}a_k - b| < \epsilon$$
Thank you very much. So I just do like this: We have $$\sum_{n=1 }^{\infty}a_{n}=\sum_{n=1 }^{m-1 }a_{n}+\sum_{n=m }^{\infty}a_{n}$$
Because $\sum_{n=1 }^{m-1 }a_{n}$ 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 $\sum_{k = m}^{n+m-1} a_k$ 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?

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:
 Originally Posted by youkito89 Because $\sum_{k = 1}^n a_n$ is a finite sum (constant) so $\sum_{n = 1}^\infty a_n$ converges â‡” $\sum_{n = m}^\infty$ converges.
I think you should be more explicit in this statement. Say $\displaystyle\sum_{n = 1}^\infty a_n = S$ What can you say about $\displaystyle\sum_{n = m}^\infty a_n$?

 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.

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

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post walter r Real Analysis 1 April 30th, 2014 12:30 PM sagicoh Real Analysis 6 December 20th, 2012 08:43 AM ducnhuandoan Number Theory 2 December 28th, 2011 11:16 PM oddlogic Calculus 11 March 28th, 2011 01:25 AM veronicak5678 Calculus 3 November 10th, 2008 02:59 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top