# Limit of series

1. Feb 2, 2016

### goldfish9776

1. The problem statement, all variables and given/known data
why we need to make x as absolute value ? as we can see, the original is x , why we we need to make x as absolute value ? is the working wrong ?

2. Relevant equations

3. The attempt at a solution

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2. Feb 2, 2016

### Simon Bridge

Look one line up from where you put the circle to see where it comes from.
What happens to the limit when x is a negative number, and you don't have the absolute value?

3. Feb 2, 2016

### goldfish9776

Ya, it come from the top. But, why would we need to put iy as absolute value?

4. Feb 2, 2016

### Simon Bridge

Because if you don't the possibility that x may be negative will give you the wrong limit. Try it and see:
What happens to the limit when x is a negative number, and you don't have the absolute value?

5. Feb 2, 2016

### PeroK

It's using the ratio test. This looks at $\lim_{k \rightarrow \infty} | \frac{a_{k+1}}{a_k}|$

The modulus should be introduced at the start, not arbitrarily half way through the calculation.

6. Feb 2, 2016

### goldfish9776

taking x=-4 , k =2 , the term is -16/27
taking x=4 , k=2 the term is 16/27 , i couldn't find out why there is a need to put modulus ???

7. Feb 2, 2016

8. Feb 2, 2016

### Simon Bridge

It's not what happens to the term, it's what happens to the limit.
PeroK has a good link (thanks).

9. Feb 2, 2016

### goldfish9776

what will happen to the limit ? i really have no idea...

10. Feb 2, 2016

### goldfish9776

well, for this part of the notes, why do we need to make r is between L and 1 , and we take r is larger than L at the lower part .... isn't the L itself is the factor r ?

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11. Feb 2, 2016

### Samy_A

No , L doesn't necessarily satisfy $| \frac{a_{k+1}}{a_k}| \leq L$ for k sufficiently large.
Any r with $L<r<1$ does satisfy $| \frac{a_{k+1}}{a_k}| \leq r$ for k sufficiently large.

As an example, take $0<L<1$, and define $(a_n)_n$ by:
$a_1=1,\ a_{n+1}=a_n(L+\frac{1}{n})$

Then $|\frac{a_{n+1}}{a_n}|=L+\frac{1}{n} > L$, but $\displaystyle \lim_{n\rightarrow +\infty} |\frac{a_{n+1}}{a_n}| =L$.

Last edited: Feb 2, 2016
12. Feb 2, 2016

### Ray Vickson

If $L = \lim_{n \to \infty} |a_{n+1}/a_n|$ then, for any (small) $\epsilon > 0$ we have $(L-\epsilon) |a_n| < |a_{n+1}| < (L+\epsilon) |a_n|$ for all $n$ sufficiently large; essentially, that is the definition of the limit above. We simply cannot conclude that $|a_{n+1}| \leq L |a_n|$, and in fact, you can construct examples where this is false.

13. Feb 2, 2016

### goldfish9776

can you give example for the explaination above ?

14. Feb 3, 2016

### Samy_A

The example in post 11 illustrates $|\frac{a_{n+1}}{a_n}|>L$ for each n, while $\displaystyle \lim_{n\rightarrow +\infty} |\frac{a_{n+1}}{a_n}| =L$.

15. Feb 4, 2016

### vela

Staff Emeritus
Maybe a picture will help. When you say $\lim_{n\to\infty}\left\lvert \frac{a_{n+1}}{a_n} \right\rvert = L$, it means when $n$ is sufficiently large, $\left\lvert \frac{a_{n+1}}{a_n} \right\rvert$ will be within $\varepsilon$ of $L$, where we can choose $\varepsilon>0$ arbitrarily. In the picture, that means the terms of the sequence will lie in the green part of the number line. While the terms are close to $L$, they could be greater than or less than $L$, so the limit doesn't guarantee that $\left\lvert \frac{a_{n+1}}{a_n} \right\rvert<L$. However, since $r>L$, there's some room between $r$ and $L$, so we can choose $\varepsilon$ such that we can guarantee that $\left\lvert \frac{a_{n+1}}{a_n} \right\rvert < r$ for $n$ sufficiently large.

16. Feb 4, 2016

### goldfish9776

what is the purpose of not making L =r , but make L less than r ? isn't the L is the factor = r ?

17. Feb 4, 2016

### vela

Staff Emeritus
A proof is an argument, so you're essentially asking, why don't you argue it this way instead. So why don't you try finishing the proof with $r$ set equal to $L$? You'll see it won't work.