text: # 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
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distilabel_metadata.content: # 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
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document is a discussion thread about a mathematical limit. It starts with a user asking why a specific limit is true, and then multiple users provide explanations and proofs. The thread contains a question and several answers, including detailed mathematical derivations and explanations.
  
  Label: question-and-answer