# If $\sum_{n=0}^\infty c_n4^n$ is convergent, is $\sum_{n=0}^\infty c_n(-4)^n$ convergent as well?

Please identify the flaw in my reasoning:

$\displaystyle \sum_{n=0}^\infty c_n4^n$ is convergent, so by the ratio test: $\displaystyle \lim_{n \to \infty}\left\vert\frac{a_{n+1}}{a_n}\right\vert = \lim_{n \to \infty}\left\vert\frac{c_{n+1}4^{n+1}}{c_n4^n}\right\vert < 1$.

$\displaystyle \lim_{n \to \infty}\left\vert\frac{c_{n+1}}{c_n}(4)\right\vert < 1 \Rightarrow \displaystyle \lim_{n \to \infty}\left\vert\frac{c_{n+1}}{c_n}\right\vert < \frac{1}{4}$

Now, applying the ratio test to $\displaystyle \sum_{n=0}^\infty c_n(-4)^n$:

$\displaystyle \lim_{n \to \infty}\left\vert\frac{a_{n+1}}{a_n}\right\vert = \lim_{n \to \infty}\left\vert\frac{c_{n+1}(-4)^{n+1}}{c_n(-4)^n}\right\vert = \lim_{n \to \infty}\left\vert\frac{c_{n+1}}{c_n}(-4)\right\vert = \lim_{n \to \infty}\left\vert\frac{c_{n+1}}{c_n}\right\vert(4)$

Since $\displaystyle \lim_{n \to \infty}\left\vert\frac{c_{n+1}}{c_n}\right\vert < \frac{1}{4}$, $\displaystyle \lim_{n \to \infty}\left\vert\frac{c_{n+1}}{c_n}\right\vert(4) < 1$.

Therefore by the ratio test, $\displaystyle \sum_{n=0}^\infty c_n(-4)^n$ is convergent.

Turns out the answer is that we cannot conclude $\displaystyle \sum_{n=0}^\infty c_n(-4)^n$ is convergent, so I'm trying to figure out where I took a wrong turn.

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If we apply the ration test to $\sum_n\frac 1{n^2}$n we don't have the conclusion you have on the first line. –  Davide Giraudo Nov 10 '12 at 18:25
Typo, I of course meant the ratio test. –  Davide Giraudo Nov 10 '12 at 18:40
Could $c_n$ itself have an alternating sign? –  GreĝRos Nov 10 '12 at 18:40
The ratio test says that if a certain limit is less than $1$, then a series converges. It does not say that if the series converges, then that limit is less than $1$. –  Michael Hardy Nov 10 '12 at 18:46
The ratio test does not tell you that $\lim |a_{n+1}/a_n| < 1$. It does tell you that this limit is not greater than $1$, but the limit could very well fall into the inconclusive region (does not exist, or is exactly equal to $1$). –  Erick Wong Dec 2 '14 at 2:02

You’re misusing the ratio test. It tells you that if the limit is less that $1$, the series converges; it does not tell you that if the series converges, the limit is less than $1$. Thus, your very first line is wrong.

Here’s an example illustrating how $\displaystyle \sum_{n=1}^\infty c_n(-4)^n$ can fail to converge even when $\displaystyle \sum_{n=1}^\infty c_n4^n$ converges.

Let $c_n=\dfrac{(-1)^n}{n4^n}$; then $$\sum_{n=1}^\infty c_n(-4)^n=\sum_{n=1}^\infty\frac{(-1)^n(-4)^n}{n4^n}=\sum_{n=1}^\infty\frac1n\;,$$

which of course diverges: it’s the harmonic series. But

$$\sum_{n=1}^\infty c_n4^n=\sum_{n=1}^\infty\frac{(-1)^n4^n}{n4^n}=\sum_{n=1}^\infty\frac{(-1)^n}n$$

is the alternating harmonic series, which is convergent.

(If you insist on having the index start at $0$, just let $c_0=0$.)

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+1 : I was about to post this same example. –  Michael Hardy Nov 10 '12 at 18:53

You can express you series as a power series $$\sum c_nx^n.$$ The radius of convergence of the series can be given by the ratio test. Suppose that radius is $r$. We have $3$ different cases

• If $r$ is $0$, then the series only converges for $x=0$.
• If $r=\infty$, then the series converges for every value of $x$
• If $0<r<\infty$, then the series converges for every values of $x$ in $(-r,r)$
• Now in this case, you need to check if it converges at $x=r,x=-r$. It is possible to converge at $-r$ but not at $r$, however if you converge at $r$, then you also converge at $-r$.
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Brian M. Scott has posted what may be among the simplest counterexamples. You can create a counterexample by starting with any conditionally convergent series and doing what he did with it.

But the question asks for identification of flaws in the reasoning. Here's one: The ratio test says that if a certain limit is less than $1$, then a series converges. It does not say that if the series converges, then that limit is less than $1$. In fact, there are some cases in which the limiting ratio is $1$ and the series converges: $\sum_n 1/n^2$ is one such case. (There are also some cases where the limiting ratio is $1$ and the series diverges. One such case is $\sum_n 1/n$. The existence of examples of both kinds is precisely the reason why it is said that when the limit is $1$ then the test is inconclusive.)

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Let

$$c_n:=\frac{(-1)^n}{n4^n}\Longrightarrow \sum_{n=1}^\infty c_n4^n=\sum_{n=1}\frac{(-1)^n}{n}\,\,\,\;\;\;\text{converges}$$

yet

$$\sum_{n=1}^\infty c_n(-4)^n=\sum_{n=1}^\infty\frac{1}{n}\,\,\,\;\;\;\;\;\text{diverges}$$

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