Find the convergence domain of the series $\sum_\limits{n=1}^{\infty}\frac{(-1)^{n-1}}{n3^n(x-5)^n}$

Question

Exercise: Find the convergence domain of the series $$\sum_\limits{n=1}^{\infty}\frac{(-1)^{n-1}}{n3^n(x-5)^n}$$

To solve the series I used the comparison test then the Leibniz criteria:

$$\sum_\limits{n=1}^{\infty}\frac{(-1)^{n-1}}{n3^n(x-5)^n}\leqslant\sum_\limits{n=1}^{\infty}\frac{(-1)^{n-1}}{(x-5)^n}$$

Now applying Leibniz criteria the series converge for $x\neq 5$.

Solution(book): $x\geqslant5\frac{1}{3},x<4\frac{2}{3}$.

Question:

The solution surprised me. What am I doing wrong? Why is it not the same?

Thanks in advance!

Answer 1

HINT

By ratio test

$$\left|\frac{(-1)^{n}}{(n+1)3^{n+1}(x-5)^{n+1}}\frac{n3^n(x-5)^n}{(-1)^{n-1}}\right|=\frac{n}{(n+1)3|x-5|}\to L<1 \iff \frac1{3|5-x|}<1$$

then consider separetely the cases $\frac1{3|5-x|}=1$.

Answer 2

The series $\sum \frac{r^n}{n}$ converges (for real $r$) iff $-1 \le r < 1$. So the series $$ \sum_\limits{n=1}^{\infty}\frac{(-1)^{n-1}}{n3^n(x-5)^n} $$ converges iff $$ -1 \le \frac{-1}{3(x-5)} < 1 $$

added

For the Liebniz criterion, we need alternating signs, and decreasing to zero in absolute value. If $$ \frac{-1}{3(x-5)} > 0 $$ we do not have alternating signs, and if $$ -1 > \frac{-1}{3(x-5)} $$ we do not have decrease to zero.

Answer 3

You're trying to use the Leibniz test, which won't tell you a specific interval of convergence (both absolute and conditional) as the question asks. It only tells you if a series happens to be conditionally convergent. Your comparison to another alternating series is also untrue; the series alternate so you can't make a generalization on it being greater or less than for all terms. For example, try plugging in $4\frac 23$ into each series:

$$\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n3^n\left(4\frac 23-5\right)^n} = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n3^n\left(-\frac 13\right)^n} = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n(-1)^n} = \sum_{n=1}^{\infty} \frac{(-1)^n}{n}$$ $$\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{\left(4\frac23-5\right)^n} = \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{\left(-\frac 13\right)^n} = -\sum_{n=1}^\infty 3^n$$

You'll see that the former is conditionally convergent (alternating harmonic) while the latter is diverging.

If you want to find the interval of convergence, then you should apply a test such as the Ratio test:

$$\require{cancel}\lim_{n\to\infty} \left| {(-1)^\cancel{{n}}\over \cancel{(n+1)}3^\cancel{n+1}(x-5)^\cancel{n+1}} \cdot {\cancel n\cancel{3^n}\cancel{(x-5)^n} \over \cancel{(-1)^{n-1}}} \right| < 1$$

So it converges for:

$$\left|-1\over 3(x-5)\right| < 1$$

First split this into two inequalities:

$${1\over 3(x-5)} < 1 \implies {1\over x-5} < 3 \implies x-5 > \frac 13 \implies x> 5\frac 13$$ $${1\over 3(x-5)} > -1 \implies {1\over x-5} > -3 \implies x-5 < -\frac 13 \implies x < 4\frac 23$$

Then check endpoints of convergence for conditional or absolute convergence, or neither. This will give you the final interval $x \geq 5\frac 13$ and $x < 4\frac 23$.

Answer 4

The OP has asked, in comments beneath various answers as well as in the question itself, what's wrong with his procedure, so I'm going to address that only.

There seem to be two fundamental misconceptions. First, in the comparison step, it looks like you are saying that

$${1\over n3^n}\le 1\implies {(-1)^{n-1}\over n3^n(x-5)^n}\le{(-1)^{n-1}\over(x-5)^n}$$

But that implication does not hold when ${(-1)^{n-1}\over(x-5)^n}$ is negative.

Second, in the "Leibniz criteria" (aka, Alternating Series Test) step, you seem to be assuming that as long as $x\not=5$, the sequence $1\over(x-5)^n$ decreases monotonically to $0$ (that's what the Alternating Series Test requires in order to conclude convegence). But this is only true if $x\gt6$. If $x\lt5$, the sequence is alternating, so it doesn't decrease, if $5\lt x\lt 6$, the sequence actually increases to $\infty$, and if $x=6$, it's the constant sequence $1$, so it doesn't decrease to $0$.

Answer 5

We have

$$ \sum_{k=0}^{\infty}(-1)^k y^k = \lim_{n\rightarrow \infty}\frac{y^n+1}{y+1} $$

here $\vert y\vert < 1 \Rightarrow$ convergence

Now

$$ \sum_{k=0}^{\infty}(-1)^k \frac{y^k}{k} = \frac{1}{y}\int_0^y \sum_{k=0}^{\infty}(-1)^k y^k dy = \frac{1}{y}\int_0^y\frac{dy}{y+1} = \frac{\ln(y+1)}{y} $$

for $y \ne 0$ and $\vert y\vert< 1$ so the convergence is linked to

$\vert\frac{1}{3(x-x)}\vert < 1 \Rightarrow 5\frac{1}{3} < x$ and $x < 4\frac{2}{3}$