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

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

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

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

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

Tonio

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

$|x_n|$

$=\ ||x_n||$

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

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

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

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

$=\ \epsilon$

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

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

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

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

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

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 $\sum _i ^\infty |x_i| < + \infty$ means $S_n = \sum\limits_{k = 1}^n {\left| {x_k } \right|} \, \Rightarrow \,\left( {\exists L} \right)\left[ {\left( {S_n } \right) \to L} \right]$.
So the sequence $S_n$ is a Cauchy sequence.
That means we can make $\left| {x_n } \right| = \left| {S_n - S_{n - 1} } \right| < \varepsilon$.
That is sufficient to prove the property: $\left| {x_n } \right| \to 0$
So proscientia's post shows the sequence is Cauchy, right?
And as you've stated the distance $|S_n - S_{n-1}| \ is <\epsilon$ but since $\epsilon >0$ 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 $|S_n - S_{n-1}| \ is <\epsilon$ but since $\epsilon >0$ 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 $S_n$ equals the limit of $S_{n-1}$ and since, as I wrote in the first message, $|x_n|=S_n-S_{n-1}$, then...

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

Tonio