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.
$\endgroup$6 Answers
$\begingroup$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.)
$\endgroup$ $\begingroup$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.
$\endgroup$ 5 $\begingroup$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.
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.
$\endgroup$ 12 $\begingroup$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 $x=-1$.
$\endgroup$ 3 $\begingroup$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.
$\endgroup$
