# Thread: Why is this limit true...?

1. ## Why is this limit true...?

I've just seen this:
$\displaystyle \{x_i\}_{i=1}^\infty \ \sum _i ^\infty |x_i| < + \infty \rightarrow lim_{+\infty} x_i = 0$
why is this true?

2. Originally Posted by bigdoggy
I've just seen this:
$\displaystyle \{x_i\}_{i=1}^\infty \ \sum _i ^\infty |x_i| < + \infty \rightarrow lim_{+\infty} x_i = 0$
why is this true?

$\displaystyle Put\,\,S:=\sum\limits_{i=1}^\infty|x_i| \Longrightarrow since\,\,|x_n|=\sum\limits_{i=1}^n |x_i|-\sum\limits_{i=1}^{n-1}|x_i|$ , passing to the limit when $\displaystyle n\rightarrow \infty$ gives the desired result.

Tonio

3. Let $\displaystyle \sum_{i\,=\,1}^\infty|x_i|=L.$ Then given $\displaystyle \epsilon>0,$ $\displaystyle \exists\,N\in\mathbb N$ such that $\displaystyle n\geqslant N\ \Rightarrow\ \left|\sum_{i\,=\,1}^n|x_i|-L\right|<\frac\epsilon2.$ Hence, $\displaystyle \forall\,n\geqslant N+1,$

$\displaystyle |x_n|$

$\displaystyle =\ ||x_n||$

$\displaystyle =\ \left|\sum_{i\,=\,1}^n|x_i|-\sum_{i\,=\,1}^{n-1}|x_i|\right|$

$\displaystyle =\ \left|\left(\sum_{i\,=\,1}^n|x_i|-L\right)-\left(\sum_{i\,=\,1}^{n-1}|x_i|-L\right)\right|$

$\displaystyle \leqslant\ \left|\sum_{i\,=\,1}^n|x_i|-L\right|+\left|\sum_{i\,=\,1}^{n-1}|x_i|-L\right|$

$\displaystyle <\ \frac\epsilon2+\frac\epsilon2$

$\displaystyle =\ \epsilon$

4. Originally Posted by tonio
$\displaystyle Put\,\,S:=\sum\limits_{i=1}^\infty|x_i| \Longrightarrow since\,\,|x_n|=\sum\limits_{i=1}^n |x_i|-\sum\limits_{i=1}^{n-1}|x_i|$ , passing to the limit when $\displaystyle n\rightarrow \infty$ gives the desired result.

Tonio
Please could you expand on this....why have you separated the $\displaystyle |x_n|$ and how do I proceed after?

5. Originally Posted by bigdoggy
Please could you expand on this....why have you separated the $\displaystyle |x_n|$ and how do I proceed after?
Originally Posted by bigdoggy
I've just seen this:
$\displaystyle \{x_i\}_{i=1}^\infty \ \sum _i ^\infty |x_i| < + \infty \rightarrow lim_{+\infty} x_i = 0$ why is this true?
To say that $\displaystyle \sum _i ^\infty |x_i| < + \infty$ means $\displaystyle S_n = \sum\limits_{k = 1}^n {\left| {x_k } \right|} \, \Rightarrow \,\left( {\exists L} \right)\left[ {\left( {S_n } \right) \to L} \right]$.
So the sequence $\displaystyle S_n$ is a Cauchy sequence.
That means we can make $\displaystyle \left| {x_n } \right| = \left| {S_n - S_{n - 1} } \right| < \varepsilon$.
That is sufficient to prove the property: $\displaystyle \left| {x_n } \right| \to 0$

6. Originally Posted by bigdoggy
Please could you expand on this....why have you separated the $\displaystyle |x_n|$ and how do I proceed after?
Do exactly as I wrote: take the limit in both sides when $\displaystyle n\rightarrow \infty$...
I represented (not separated) $\displaystyle x_n$ as I did above so that we'll be able to evaluate the sequence's limit.

Tonio

7. Originally Posted by tonio
Do exactly as I wrote: take the limit in both sides when $\displaystyle n\rightarrow \infty$...
I represented (not separated) $\displaystyle x_n$ as I did above so that we'll be able to evaluate the sequence's limit.

Tonio
So, evaluate $\displaystyle lim_{n \rightarrow \infty} \sum ^n _1x_i - lim_{n \rightarrow \infty} \sum^{n-1} _1 x_i$

Am I ok to drop the modulus' when evaluating the limit as above?
I'm not sure how to evaluate this

8. Originally Posted by Plato
To say that $\displaystyle \sum _i ^\infty |x_i| < + \infty$ means $\displaystyle S_n = \sum\limits_{k = 1}^n {\left| {x_k } \right|} \, \Rightarrow \,\left( {\exists L} \right)\left[ {\left( {S_n } \right) \to L} \right]$.
So the sequence $\displaystyle S_n$ is a Cauchy sequence.
That means we can make $\displaystyle \left| {x_n } \right| = \left| {S_n - S_{n - 1} } \right| < \varepsilon$.
That is sufficient to prove the property: $\displaystyle \left| {x_n } \right| \to 0$
So proscientia's post shows the sequence is Cauchy, right?
And as you've stated the distance $\displaystyle |S_n - S_{n-1}| \ is <\epsilon$ but since $\displaystyle \epsilon >0$ as in a Cauchy sequence, that implies it must tend to zero....is my thinking right?

9. Originally Posted by bigdoggy
So proscientia's post shows the sequence is Cauchy, right?
And as you've stated the distance $\displaystyle |S_n - S_{n-1}| \ is <\epsilon$ but since $\displaystyle \epsilon >0$ as in a Cauchy sequence, that implies it must tend to zero....is my thinking right?

No, but close: since the sequence (of partial sums of the series) is Cauchy then it has a limit, but since the limit of $\displaystyle S_n$ equals the limit of $\displaystyle S_{n-1}$ and since, as I wrote in the first message, $\displaystyle |x_n|=S_n-S_{n-1}$, then...

Be careful: $\displaystyle S=\lim_{n\rightarrow \infty}S_n$ can NOT be zero unless $\displaystyle x_i=0\,\,\forall\,i$ !

Tonio
The content of the text is a discussion about a mathematical limit and its proof, involving a series and the concept of a Cauchy sequence.
multiple_questions_with_answers
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
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- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
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# Thread: Why is this limit true...?

1. ## Why is this limit true...?

I've just seen this:
$\displaystyle \{x_i\}_{i=1}^\infty \ \sum _i ^\infty |x_i| < + \infty \rightarrow lim_{+\infty} x_i = 0$
why is this true?

2. Originally Posted by bigdoggy
I've just seen this:
$\displaystyle \{x_i\}_{i=1}^\infty \ \sum _i ^\infty |x_i| < + \infty \rightarrow lim_{+\infty} x_i = 0$
why is this true?

$\displaystyle Put\,\,S:=\sum\limits_{i=1}^\infty|x_i| \Longrightarrow since\,\,|x_n|=\sum\limits_{i=1}^n |x_i|-\sum\limits_{i=1}^{n-1}|x_i|$ , passing to the limit when $\displaystyle n\rightarrow \infty$ gives the desired result.

Tonio

3. Let $\displaystyle \sum_{i\,=\,1}^\infty|x_i|=L.$ Then given $\displaystyle \epsilon>0,$ $\displaystyle \exists\,N\in\mathbb N$ such that $\displaystyle n\geqslant N\ \Rightarrow\ \left|\sum_{i\,=\,1}^n|x_i|-L\right|<\frac\epsilon2.$ Hence, $\displaystyle \forall\,n\geqslant N+1,$

$\displaystyle |x_n|$

$\displaystyle =\ ||x_n||$

$\displaystyle =\ \left|\sum_{i\,=\,1}^n|x_i|-\sum_{i\,=\,1}^{n-1}|x_i|\right|$

$\displaystyle =\ \left|\left(\sum_{i\,=\,1}^n|x_i|-L\right)-\left(\sum_{i\,=\,1}^{n-1}|x_i|-L\right)\right|$

$\displaystyle \leqslant\ \left|\sum_{i\,=\,1}^n|x_i|-L\right|+\left|\sum_{i\,=\,1}^{n-1}|x_i|-L\right|$

$\displaystyle <\ \frac\epsilon2+\frac\epsilon2$

$\displaystyle =\ \epsilon$

4. Originally Posted by tonio
$\displaystyle Put\,\,S:=\sum\limits_{i=1}^\infty|x_i| \Longrightarrow since\,\,|x_n|=\sum\limits_{i=1}^n |x_i|-\sum\limits_{i=1}^{n-1}|x_i|$ , passing to the limit when $\displaystyle n\rightarrow \infty$ gives the desired result.

Tonio
Please could you expand on this....why have you separated the $\displaystyle |x_n|$ and how do I proceed after?

5. Originally Posted by bigdoggy
Please could you expand on this....why have you separated the $\displaystyle |x_n|$ and how do I proceed after?
Originally Posted by bigdoggy
I've just seen this:
$\displaystyle \{x_i\}_{i=1}^\infty \ \sum _i ^\infty |x_i| < + \infty \rightarrow lim_{+\infty} x_i = 0$ why is this true?
To say that $\displaystyle \sum _i ^\infty |x_i| < + \infty$ means $\displaystyle S_n = \sum\limits_{k = 1}^n {\left| {x_k } \right|} \, \Rightarrow \,\left( {\exists L} \right)\left[ {\left( {S_n } \right) \to L} \right]$.
So the sequence $\displaystyle S_n$ is a Cauchy sequence.
That means we can make $\displaystyle \left| {x_n } \right| = \left| {S_n - S_{n - 1} } \right| < \varepsilon$.
That is sufficient to prove the property: $\displaystyle \left| {x_n } \right| \to 0$

6. Originally Posted by bigdoggy
Please could you expand on this....why have you separated the $\displaystyle |x_n|$ and how do I proceed after?
Do exactly as I wrote: take the limit in both sides when $\displaystyle n\rightarrow \infty$...
I represented (not separated) $\displaystyle x_n$ as I did above so that we'll be able to evaluate the sequence's limit.

Tonio

7. Originally Posted by tonio
Do exactly as I wrote: take the limit in both sides when $\displaystyle n\rightarrow \infty$...
I represented (not separated) $\displaystyle x_n$ as I did above so that we'll be able to evaluate the sequence's limit.

Tonio
So, evaluate $\displaystyle lim_{n \rightarrow \infty} \sum ^n _1x_i - lim_{n \rightarrow \infty} \sum^{n-1} _1 x_i$

Am I ok to drop the modulus' when evaluating the limit as above?
I'm not sure how to evaluate this

8. Originally Posted by Plato
To say that $\displaystyle \sum _i ^\infty |x_i| < + \infty$ means $\displaystyle S_n = \sum\limits_{k = 1}^n {\left| {x_k } \right|} \, \Rightarrow \,\left( {\exists L} \right)\left[ {\left( {S_n } \right) \to L} \right]$.
So the sequence $\displaystyle S_n$ is a Cauchy sequence.
That means we can make $\displaystyle \left| {x_n } \right| = \left| {S_n - S_{n - 1} } \right| < \varepsilon$.
That is sufficient to prove the property: $\displaystyle \left| {x_n } \right| \to 0$
So proscientia's post shows the sequence is Cauchy, right?
And as you've stated the distance $\displaystyle |S_n - S_{n-1}| \ is <\epsilon$ but since $\displaystyle \epsilon >0$ as in a Cauchy sequence, that implies it must tend to zero....is my thinking right?

9. Originally Posted by bigdoggy
So proscientia's post shows the sequence is Cauchy, right?
And as you've stated the distance $\displaystyle |S_n - S_{n-1}| \ is <\epsilon$ but since $\displaystyle \epsilon >0$ as in a Cauchy sequence, that implies it must tend to zero....is my thinking right?

No, but close: since the sequence (of partial sums of the series) is Cauchy then it has a limit, but since the limit of $\displaystyle S_n$ equals the limit of $\displaystyle S_{n-1}$ and since, as I wrote in the first message, $\displaystyle |x_n|=S_n-S_{n-1}$, then...

Be careful: $\displaystyle S=\lim_{n\rightarrow \infty}S_n$ can NOT be zero unless $\displaystyle x_i=0\,\,\forall\,i$ !

Tonio
user
The content of the text is a discussion about a mathematical limit and its proof, involving a series and the concept of a Cauchy sequence.

TYPE: MULTIPLE_QUESTIONS_WITH_ANSWERS