# Why is the ratio test not working for this series?

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!

• 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? – RRR Feb 16 '20 at 19:40
• 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. – bjorn93 Feb 16 '20 at 21:18
• The ratio test applies to any series. – DanielWainfleet Feb 16 '20 at 23:02

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.

• Does the ratio test not necessarily give the best bounds then? Because it converges whenever $|x|>1$. – RRR Feb 16 '20 at 19:42
• 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 $\sqrt{2} > 1$. – Nicholas Roberts Feb 16 '20 at 19:56
• Most of books written by human been contain errors. – hamam_Abdallah Feb 16 '20 at 19:59

Definition of the Ratio Test taken from Gilbert Strang's "Calculus" testbook: 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 $$\left | x \right |>1$$.

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.

• 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. – Nicholas Roberts Feb 17 '20 at 1:03
• @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 $2$. – bjorn93 Feb 17 '20 at 1:30
• Thanks for the reply. So what were you regarding to be $a_n$ in your answer? Was it $\frac{2}{1+x^2^n}$? – Nicholas Roberts Feb 17 '20 at 1:35
• @NicholasRoberts $2^n/\left(1+x^{2n}\right)$ – bjorn93 Feb 17 '20 at 1:37

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.