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For what values of x does the series $1+\frac{x}{3}+\frac{x^2}{5}+\frac{x^3}{7}+\cdot\cdot\cdot$ converge?

The solution states:
The general term is of the form $u_n(x)=\frac{x^{n-1}}{(2n-1)}$, and hence $$\frac{|u_{n+1}|}{|u_n|}=\frac{|x^n|}{(2n+1)}\cdot\frac{(2n-1)}{|x^{n-1}|}$$ ------edit start------- $$=\frac{(2n-1)}{(2n+1)}\cdot\frac{|x^n|}{|x^{n-1}|}$$ ------ edit end ------- $$=\frac{(2n-1)}{(2n+1)}|x|$$ clearly

My question is:

  1. How do you get from $1+\frac{x}{3}+\frac{x^2}{5}+\frac{x^3}{7}+\cdot\cdot\cdot$ to $u_n(x)=\frac{x^{n-1}}{(2n-1)}$?
  2. Why do the absolute restriction only apply to the $|x^n|$ and $|x^{n-1}|$ in the next line and not to the rest of the equation?
  3. and lastly, We go from $=\frac{(2n-1)}{(2n+1)}\cdot\frac{|x^n|}{|x^{n-1}|}$ to $=\frac{(2n-1)}{(2n+1)}|x|$ I.E. how does $\frac{|x^n|}{|x^{n-1}|}=|x|$ in the next line? (I understand that once you apply the limits, that $\frac{(2n-1)}{(2n+1)} = 1$)

I have added a line to the equation that was not there earlier and clarified my last question.

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For 1: it's one of those patterns you notice; you see the odd numbers in the denominator, and the nonnegative integers as powers, so... – J. M. May 2 '12 at 6:27
For 2: the thing that was taken out of the absolute value sign is alright, since it's always positive for $n > 0$. – J. M. May 2 '12 at 6:28
For 3: $$\left|\frac{a}{b}\right|=\frac{|a|}{|b|}$$ – J. M. May 2 '12 at 6:29
up vote 2 down vote accepted
  1. They are defining the terms of your sum. $u_1(x)=\frac{x^{1-1}}{2-1}=1$ and $u_2(x)=\frac{x^{2-1}}{4-1}=\frac{x}{3}$ and $u_3(x)=\frac{x^{3-1}}{6-1}=\frac{x^2}{5}$ and so on and so forth.

2.$|u_n|=\left|\frac{x^n}{2n+1}\right|$ but since $2n+1>0$ because we are dealing in positive integers then we can remove the absolute value around them resulting in $|u_n|=\frac{|x^n|}{2n+1}$ similar reasoning for $u_{n+1}$

3.basic exponent rules for division and a little bit of absolute value rules thrown in $\frac{|x^n|}{|x^{n-1}|}=\frac{|x|^n}{|x|^{n-1}}=|x|$

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On point 3: Aaaaa, so: $$\frac{x^n}{x^{n-1}} = \frac{x^n}{x^n \cdot x^{-1}}$$ So, the two $x^n$'s cancel out leaving $$=\frac{1}{x^{-1}}$$ $$=x$$ Thanks. Highschool math :-\. – Gineer May 2 '12 at 17:00

Let me deal with your second and third questions first, as they're easier. For (2), there is no need to put absolute value signs around $2n-1$ or $2n+1$, because both of these expressions are non-negative already: $n\ge 1$. For (3), the line in question does not say that $\frac{|x^n|}{|x^{n-1}|}=|x|$: it says $$\lim\limits_{n\rightarrow\infty}\frac{|u_{n+1}|}{|u_n|}=|x|\;.$$ Note the limit: it's essential! This can be expanded as follows: $$\lim\limits_{n\rightarrow\infty}\frac{|u_{n+1}|}{|u_n|}=\lim_{n\to\infty}\frac{2n-1}{2n+1}|x|=|x|\lim_{n\to\infty}\frac{2n-1}{2n+1}=|x|\cdot 1=|x|\;,$$ since $|x|$ is a constant with respect to the limit over $n$.

Now let's look at your first question. You have the series $1+\frac{x}{3}+\frac{x^2}{5}+\frac{x^3}{7}+\dots$, and you want to know how to determine that the $n$-th term, $u_n(x)$, is $\frac{x^{n-1}}{2n-1}$. First, note that whether or not you could have worked this out yourself, you can easily verify that it's correct. The formula gives $\frac{x^{1-1}}{2\cdot1-1}=\frac11=1$ for the first term, which is correct. Each time you increase $n$ by $1$ in the formula for $u_n(x)$, the exponent in the numerator increases by $1$, and the denominator increases by $2$; since that's exactly what's happening in the series $1+\frac{x}{3}+\frac{x^2}{5}+\frac{x^3}{7}+\dots$, and since the formula gets the first term right, it must get every term right.

How would you come up with the formula yourself? You might notice that the exponents in the numerator are increasing by $1$ each term; that suggests something like $x^n$. However, the exponent in the first term is $0$, that in the second term is $1$, and so on: the exponent is always one less than the term number, so the numerator of the $n$-th is actually $x^{n-1}$. The denominators are $1,3,5,7,\dots$, increasing by $2$ each time. This suggests that the denominator is something like $2n$, which increases by $2$ every time $n$ is increased by $1$. But this would make the denominators $2,4,6,8,\dots$, exactly one more than we want, so we subtract one: the denominator of the $n$-th term is $2n-1$. And that gives us the formula: the $n$-th term is $\frac{x^{n-1}}{2n-1}$.

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You can see the pattern, in the numerator It's: $$x^0, x^1, x^2, x^3, \dots, x^{n-1}$$ And in the denominator, it's the odd numbers: $$1, 3, 5, 7, 9, \dots, 2n-1$$

Therefore, it is:


While all odd numbers are positive ($n>0$), you can take them out of the absolute and $2n+1$ produce the same series of number as $2n-1$ which is odd numbers:

$$\frac{|u_{n+1}|}{|u_n|}=\frac{|x^n|}{\color{red}{(2n+1)}}\cdot\frac{\color{red}{(2n-1)}}{|x^{n-1}|} = \frac{|x^n|}{|x^{n-1}|}=|\frac{x^n}{x^{n-1}}|=|x|$$

Hence: $$\lim\limits_{n\rightarrow\infty}\frac{|u_{n+1}|}{|u_n|}=|x|$$

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for $2n-1, n>0.$ $n_1 = 1$ so $2(1)-1 = 1$, $n_2 = 2$ is $3$, $n_3 = 3$ is $5$ and $n_4 = 2$ is $7$ however for $2n+1, n>0$. $n_1 = 1$ so $2(1)+1 = 3$, $n_2 = 2$ is $5$, $n_3 = 3$ is $7$ and $n_4 = 2$ is $9$ So although they both create a series of odd numbers, they are not equal for any given $n$ in the series. What am I misunderstanding? – Gineer May 2 '12 at 16:50
Once the limits are applied I agree that $\frac{(2n-1)}{(2n+1)}=1$ since $[2(\infty)-1] = [2(\infty)+1]$ – Gineer May 2 '12 at 17:05
@Gineer: They produce the same series of numbers, one staring at $n=1$, the other at $n=0$. – Gigili May 2 '12 at 18:34
Also, @Gineer, $|\frac{x^n}{x^{n-1}}|= |\frac{x \cdot x^{n-1}}{x^{n-1}}|$. – Gigili May 2 '12 at 18:36

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