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For which values of $x$ does the series presented below converge?

$$\sum_{n=1}^{+\infty}\frac{x^n(1-x^n)}{n}$$

Neither the root test nor the ratio test is of much help - I've tried for awhile now - so any hints would be greatly appreciated.

Thanks.

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To apply the ratio test we have to consider $$r_n=\frac{x^{n+1}(1-x^{n+1})}{n+1}\frac{n}{x^n(1-x^n)}=x\frac{1-x^{n+1}}{1-x^n}\ .$$

  • If $|x|<1$ then $r_n\to x$ and the series converges.
  • If $|x|>1$ then $$r_n=x^2\frac{1-x^{-(n+2)}}{1-x^{-n}}\to x^2$$ and the series diverges.
  • If $x=1$ then the series vanishes (and converges).
  • If $x=-1$ the series is $$-2\Bigl(\frac11+\frac13+\frac15+\cdots\Bigr)$$ which diverges.
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  • $\begingroup$ (the x=-1 case was tricky to handle ... but I agree with your expansion) $\endgroup$ – user258002 Jul 31 '15 at 2:29
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The ratio test does work here:

$$\begin{align} \left|\frac{a_{n+1}}{a_n}\right| &= \left|\frac{x^{n+1}(1-x^{n+1})}{x^n(1-x^n)}\right|\cdot\frac{n}{n+1} \\ &= \left|x\cdot\frac{1-x^{n+1}}{1-x^n}\right|\cdot\frac{n}{n+1} \end{align}$$

If $|x|<1$ then as $n\to\infty$, $x^n\to 0$ and that ratio tends to $x$.

If $|x|=1$ then the series obviously converges for $x=1$ and diverges for $x=-1$.

If $|x|>1$ then the ratio can be written as

$$\left|x\cdot\frac{\frac 1{x^n+1}-1}{\frac 1{x^n+1}-\frac 1x}\right|\cdot\frac{n}{n+1}$$

and that approaches $x\cdot\frac{0-1}{0-\frac 1x}=x^2$.

In all those cases, the radius of convergence is $1$, including convergence at $x=1$ but not $x=-1$.

(Yes, I know we could have ignored the case $|x|>1$, but I wanted to be consistent here in the approach.)

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If $|x|>1$ so $|x^n(1-x^n)|/n$ does not converge to $0$, and if $|x|<1$ we can take the sum separeted.

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  • $\begingroup$ Or just note that $|x^n(1-x^n)|<2|x|^n$ when $|x|<1$ - you don't need to separate. $\endgroup$ – Thomas Andrews Jul 31 '15 at 1:14
  • $\begingroup$ $x=1$ implies sum 0 $\endgroup$ – Euler88 ... Jul 31 '15 at 1:16
  • $\begingroup$ Not clear what your point is with that comment. $\endgroup$ – Thomas Andrews Jul 31 '15 at 1:17
  • $\begingroup$ It's only a complement for my answer.. You're right with respect your comment $\endgroup$ – Euler88 ... Jul 31 '15 at 1:22
  • $\begingroup$ You can edit your answer to put that information in it. It also does not converge when $x=-1$. Comments should be used to "update" or amend an answer. $\endgroup$ – Thomas Andrews Jul 31 '15 at 1:30
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Split it into two series $$ \sum_n\frac{x^n}{n}-\sum_n\frac{x^{2n}}{n} $$ Both of these series are individually amenable to the ratio test.

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  • 1
    $\begingroup$ But that doesn't mean it isn't possible for the first series to converge for more $x$. It is true that if $|x|<1$ you can conclude the first series converges since these two converge, but you can't conclude that it doesn't converge for $|x|>1$. A simple example is $x=1$ - the first series trivially converges then, but neither of your two series converges. $\endgroup$ – Thomas Andrews Jul 31 '15 at 1:02
  • $\begingroup$ @ThomasAndrews That is in general true, but for this specific example you will correctly find the radius of convergence. I was only trying to give a general hint, but I will edit to make things more specific. $\endgroup$ – Jahan Claes Jul 31 '15 at 1:05
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    $\begingroup$ Well, it is definitely true in this case - for $x=1$. The OP didn't ask for a radius of convergence, OP asked for all $x$ for which convergence. But I agree that it does not converge when $|x|>1$, but it is something that has to be proved. $\endgroup$ – Thomas Andrews Jul 31 '15 at 1:06
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    $\begingroup$ Except this sort of series doesn't necessarily have a radius of convergence. $\endgroup$ – Thomas Andrews Jul 31 '15 at 1:08
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    $\begingroup$ It will once I've shown it does. $\endgroup$ – Jahan Claes Jul 31 '15 at 1:09
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CASE 1: $|x|<1$

We recall that the series for $-\log (1-x)$ is

$$-\log (1-x)=\sum_{n=1}^{\infty}\frac{x^n}{n}$$

for $|x|<1$. For $|x|<1$, the series of interest is

$$\begin{align} \sum_{n=1}^{\infty}\frac{x^n(1-x^n)}{n}&=\sum_{n=1}^{\infty}\frac{x^n}{n}-\sum_{n=1}^{\infty}\frac{(x^2)^n}{n}\\\\ &=-\log(1-x)+\log(1-x^2)\\\\ &=\log(1+x) \end{align}$$


CASE 2: $|x|=1$

To test the convergence at $x=\pm 1$, we note that for $x=1$ the series is trivial (i.e, equal to zero). For $x=-1$,we have

$$\sum_{n=1}^{\infty}\frac{(-1)^n(1-(-1)^n)}{n}=-\sum_{n=1}^{\infty}\frac{2}{2n+1}$$

which clearly diverges.


CASE 3: $|x|>1$

For $x>1$ we observe that

$$|x^n-x^{2n}|> \frac12 x^{2n}$$

for $n$ sufficiently large (for any fixed $|x|>1$, take $n>(\log 2)/\log |x|$) and therefore the series diverges for $|x|>1$.


Putting everything together we have

$$\sum_{n=1}^{\infty}\frac{x^n(1-x^n)}{n}= \begin{cases} -\log (1+x)&,|x|<1\\\\ 0&,x=1 \end{cases}$$

and $\sum_{n=1}^{\infty}\frac{x^n(1-x^n)}{n}$ diverges for $|x|>1$ and $x=-1$.

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  • $\begingroup$ This doesn't come even close to answering the question. $\endgroup$ – Thomas Andrews Jul 31 '15 at 1:13
  • $\begingroup$ @ThomasAndrews Yeah, I edited it ... any other comments are welcome! $\endgroup$ – Mark Viola Jul 31 '15 at 1:15
  • $\begingroup$ @user258002 Please let me know how I can improve my answer. I really just want to give you the best answer I can. $\endgroup$ – Mark Viola Aug 6 '15 at 0:28
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Use the ratio test:

$\displaystyle \lim_{n \to \infty} \frac{\frac{x^{n+1}(1-x^{n+1})}{n+1}}{\frac{x^n(1-x^n)}{n}} = \lim_{n \to \infty} \frac{n (x-x^{n+2})}{(n+1) (1-x^{n})} = \lim_{n \to \infty} \frac{ x-x^{n+2}}{ 1-x^{n}} = \lim_{n \to \infty} \frac{ x ( 1-x^{n+1})}{ 1-x^{n}}$

We now see that the numerator is similar to $\ x$ times the denominator.

Consider the cases: $\ |x| > 1$, $\ |x| = 1$, $\ |x| < 1$.

$\ |x| > 1$: Then the dominant term in the limit is $\ \frac{x^{n+2}}{ x^{n}}$, which clearly goes to infinity. This doesn't converge.

$\ |x| = 1$: If x = 1, then it converges by L'Hopitals. If x = -1, the the numerator oscillates between 0 and 2. This doesn't converge.

$\ |x| < 1$: $\ x^{n+1}$ and $\ x^{n}$ both converge to 0. The dominant term is now $\ \frac{x}{1}$. This converges.

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