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Show that the series $\frac{1}{1+x} + \frac{2}{1+x^2} + ... + \frac{2^n}{1+x^{2n}} + ...$ converges when $|x|>1$, and find it's sum.

I tried using the ratio test, but I first rewrote the above series as $\sum_{n=0}^{\infty}\frac{2^n(1-x^{2^n})}{1-x^{2^{n+1}}}$. The ratio test on this series gets us $\lim_{n\to\infty} \frac{2}{x^2}$ after disregarding the terms that disappear and computing it. However, in this case, $|x|>\sqrt2$, which is incorrect.

Where did I go wrong? I'm pretty sure I rewrote the series correctly, so this should give the correct answer. The solution is apparently to find a formula after testing the first few terms and proving by induction, but I am much more curious to why my method did not work, as that may show me some pitfalls that I am falling into when I am using the ratio test.

Thank you!

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  • $\begingroup$ I mean, a ratio test is just a comparison to a geometric series, so I don't know why it only applies to power series? $\endgroup$ – RRR Feb 16 '20 at 19:40
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    $\begingroup$ Is the term of the series $2^n/\left(1+x^{2n}\right)$ or is it $2^n/\left(1+x^{2^n}\right)$? Seems like there is a discrepancy between what you wrote in the first $2$ lines. $\endgroup$ – bjorn93 Feb 16 '20 at 21:18
  • $\begingroup$ The ratio test applies to any series. $\endgroup$ – DanielWainfleet Feb 16 '20 at 23:02
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The ratio test gives

$$\lim_{n\to\infty}\frac{2}{x^2}$$

the series converges if $$|x|>\sqrt{2}.$$

If $|x|=\sqrt{2},$ the general term becomes

$$\frac{2^n}{1+2^n}$$ which goes to $1$, the series diverges.

If $|x|<\sqrt{2}$, it diverges.

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  • $\begingroup$ Does the ratio test not necessarily give the best bounds then? Because it converges whenever $|x|>1$. $\endgroup$ – RRR Feb 16 '20 at 19:42
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    $\begingroup$ It does not converge for $|x| > 1$ since as Hamam pointed out, when you plug in $\sqrt{2}$, the term tends to 1. So by the divergence test, it diverges. And $\sqrt{2} > 1$. $\endgroup$ – Nicholas Roberts Feb 16 '20 at 19:56
  • $\begingroup$ Most of books written by human been contain errors. $\endgroup$ – hamam_Abdallah Feb 16 '20 at 19:59
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Definition of the Ratio Test taken from Gilbert Strang's "Calculus" testbook:

enter image description here

Apply the "Ratio Test" to the problem: $$\frac{a_{n+1}}{a_{n}}=\frac{2^{n+1}}{1+x^{2n+1}}\cdot \frac{1+x^{2n}}{2^{n}}=\frac{2\left ( 1+x^{2n} \right )}{1+x^{2n+1}}$$

$$\lim_{n\rightarrow \infty }\frac{2\left ( 1+x^{2n} \right )}{1+x^{2n+1}}=2\lim_{n\rightarrow \infty }\frac{ 1+x^{2n}}{1+x^{2n+1}}=2\lim_{n\rightarrow \infty }\frac{1/x^{2n}+1}{1/x^{2n}+x}......\left(1\right)$$

$$\left | x \right |>1$$

Expression (1) = $2\lim_{n\rightarrow \infty }\frac{1}{x}=\frac{2}{x}$

The only thing that is certain about $\frac{2}{x}$ is that it is smaller than $2$, since $\left | x \right |>1$. There is no guarantee that it is smaller than $1$. In other words, the ratio test failed.

Try the "Root test":

$$\sqrt{\frac{x^{2n}}{1+x^{2n}}}< \frac{x^n}{1+x^{n}}$$

$$\lim_{n\rightarrow \infty }\frac{x^{n}}{1+x^{n}}=\lim_{n\rightarrow \infty }\frac{1}{1/x^{n}+1}......\left ( 2 \right )$$

$\left | x \right |> 1$ The limit in Expression (2) is $1$

That means $\lim_{n\rightarrow \infty }\sqrt{\frac{x^{2n}}{1+x^{2n}}}< 1$ if $\left | x \right |>1$ By the Root Test, the series converges if $\left | x \right |>1$.

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Let $a_n$ be the term of the series. For the series to converge, $$\lim_{n\to\infty}a_n=\lim_{n\to\infty}\frac{2^n}{1+x^{2n}}=\lim_{n\to\infty}\frac{1}{\frac{1}{2^n}+\left(\frac{x^2}{2}\right)^n}=0 $$ has to be true. That's a necessary condition. This implies that $x^2/2>1\Leftrightarrow |x|>\sqrt{2}$ (think about $\lim_{n\to\infty}q^n$ for $|q|<1, |q|=1$, and $|q|>1$). The ratio test shows that $|x|>\sqrt{2}$ is also a sufficient condition: $$\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=\frac{2}{x^2}<1\,\,\text{for }|x|>\sqrt{2} $$ Hence, the series converges iff $|x|>\sqrt{2}$. And the ratio test works here.

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  • $\begingroup$ Can you explain how the ratio $\frac{a_{n+1}}{a_n}$ simplifies to $\frac{2}{x^2}$? For some reason, I am unable to get this. $\endgroup$ – Nicholas Roberts Feb 17 '20 at 1:03
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    $\begingroup$ @NicholasRoberts Note that OP might have copied the problem wrong, so it's not clear what the problem is. Otherwise, $$\frac{a_{n+1}}{a_n}=2\frac{1+x^{2n}}{1+x^{2n+2}}=2\frac{1/x^{2n}+1}{1/x^{2n}+x^2} $$ then take limit assuming that $|x|>1\Rightarrow 1/x^{2n}\to 0$. If $|x|<1$, the limit is just $2$. $\endgroup$ – bjorn93 Feb 17 '20 at 1:30
  • $\begingroup$ Thanks for the reply. So what were you regarding to be $a_n$ in your answer? Was it $\frac{2}{1+x^2^n}$? $\endgroup$ – Nicholas Roberts Feb 17 '20 at 1:35
  • $\begingroup$ @NicholasRoberts $2^n/\left(1+x^{2n}\right)$ $\endgroup$ – bjorn93 Feb 17 '20 at 1:37
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A quick check:

$\dfrac{2^n}{1+x^{2n}} \lt (\dfrac{2}{x^2})^n=:a_n$.

Ratio test, or recalling criteria for geometric series:

1) $x^2 >2$, $\sum a_n$ converges

By comparison test the original series converges for $x^2>2$.

2) Let $x^2\le 2$:

$\dfrac{2^n}{1+x^{2n}} \ge\dfrac{2^n}{1+2^n}:=b_n$.

$\lim_{n \rightarrow \infty} b_n=1 \not =0$, original series diverges.

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