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January 2nd, 2016, 08:54 AM   #1
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Question $\sum_{n=1 }^{+∞ }a_{n}$ and $\sum_{n=m }^{+∞ }a_{n}$ 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 $$\sum_{n=1}^{+\infty }a_{n}=\alpha + \sum_{n=m}^{+\infty }a_{n}$$
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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.
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January 2nd, 2016, 01:38 PM   #2
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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
$$|t_n + \sum_{k = 1}^{m - 1}a_k - b| < \epsilon$$
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January 2nd, 2016, 07:21 PM   #3
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Quote:
Originally Posted by Azzajazz View Post
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?
I'm so confused about this.
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January 3rd, 2016, 01:31 AM   #4
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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 ⇔ $\sum_{n = m}^\infty$ 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 $\displaystyle\sum_{n = m}^\infty a_n$?
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January 3rd, 2016, 02:11 AM   #5
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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.
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January 3rd, 2016, 12:58 PM   #6
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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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