How would you manipulate this series to use the alternating series test?

Question

$$\sum_{n=1}^\infty \frac{(-1)^{n}n^{1/3}}{n}$$

In the theorem I have been taught,

$a_n$ must be a decreasing sequence of positive terms such that $a_n$ tends to 0 as n tends to infinity.

Then $\sum_{n=1}^\infty (-1)^{n+1} a_n$ is convergent.

I can see that multiplying the original series by $\frac{-1}{-1}$will make the power correct for the conditions, but then when splitting the series into the defined part and $a_n$, $a_n$ is no longer a sequence of positive terms? Not sure where to go from here or if I’ve gone about this the right way initially?

Answer 1

The problem that you are having (and which is not being addressed by either of the other answers) is that you are attempting to apply a formula without understanding what that formula actually says or where it comes from. So, let's try to understand what the theorem actually says, beginning with a statement of the theorem (note: here, I really do mean the indeterminate article "a", and not "the"; as I'll discuss below, there are more general statements):

Theorem (Alternating Series Test, v1): Let $(a_0, a_1, a_2, \dotsc)$ be a sequence of real numbers such that

1. $a_n \ge 0$ for all natural numbers $n$,
2. $a_{n+1} \le a_{n}$ for all natural numbers $n$, and
3. $\lim_{n\to \infty} a_n = 0$.

Then the series $$ \sum_{n=0}^{\infty} (-1)^n a_n $$ converges.

The basic idea of a proof is as follows: let $S_k$ denote the $k$-th partial sum, i.e. $$ S_k := \sum_{n=0}^{k} (-1)^n a_n. $$ Then the sequence of even partial sums $(S_0, S_2, S_4, \dotsc)$ is a decreasing sequence, since $$ S_{2(k+1)} = S_{2k} + a_{2k+2} - a_{2k+1} \le S_{2k}, $$ which follows from the hypotheses that the $a_n$ are all nonnegative and that $a_{n+1} \le a_{n}$ for all $n$. By similar reasoning, the sequence of odd partial sums is increasing. Combining these $$ S_1 \le S_{2k+1} = S_{2k} - a_{k+1} \le S_{2k} \le S_{2}, $$ where the first inequality follows from the fact that the odd partial sums are increasing, the middle inequality follows from the fact that $a_{k+1}$ is positive, and the last inequality follows from the fact that the even partial sums are decreasing. But then the two sequences (of even and odd partial sums) must converge: they are each monotone and bounded. To see that the converge to the same thing, note that $$ \lim_{k\to\infty} |S_{k+1} - S_k| = \lim_{k\to\infty} |a_{k+1}| = 0. $$

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Now, proofs are often hard for beginners to really grasp, so here is an intuition: we alternate adding and subtracting terms, but the terms we add or subtract get smaller and smaller. This means that the partial sums oscillate (i.e. they jump up and down as we add terms), but that the oscillations get smaller and smaller. The partial sums are bouncing up and down in a smaller and smaller range, which eventually tightens down to a limit.

This implies a couple of things:

• We don't really need the terms to all be positive. Instead, we could restate the first hypothesis as "the odd and even terms of the sequence $(a_0,a_1,a_2,\dotsc)$ must have different signs." If we do that, the conclusion of the theorem becomes $$ \sum_{n=0}^{\infty} a_n $$ exists. Note that we no longer have the factor of $(-1)^n$, since we are assuming that the terms themselves alternate.
• Even if we don't want to drop the hypothesis that all of the terms have the same sign, we could instead assume that they are all negative. That is, if we assume that $a_n \le 0$ and that $a_{n+1} \ge a_n$ for all $n$, then the conclusion of the theorem is the same, i.e. $$ \sum_{n=0}^{\infty} (-1)^{n} a_n $$ exists.
• If all of the $a_n$ have the same sign, then we can multiply by $-1$ (twice) to get $$ L = \sum_{n=0}^{\infty} (-1)^{n} a_n = (-1)(-1) \sum_{n=0}^{\infty} (-1)^{n} a_n = (-1) \sum_{n=0}^{\infty} (-1)(-1)^{n} a_n = -\sum_{n=0}^{\infty} (-1)^{n+1} a_n. $$ Thus we can replace the exponent $n$ with $n+1$ without altering the convergence of the series.
• One other thought: when working with series, it is important to remember that the first finite collection of terms don't matter with respect to convergence. The series $$ \sum_{n=0}^{\infty} a_n $$ converges if and only if the series $$ \sum_{n=N}^{\infty} a_n $$ for any natural number $N$. When dealing with infinitary processes (such as infinite series), we can generally ignore anything which happens finitely often.

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With the above comments in mind, I might restate the theorem as follows:

Theorem (Alternating Series Test, v2): Let $(a_0, a_1, a_2, \dotsc)$ be a sequence such that

1. the terms eventually alternate signs, i.e. there is an $N$ sufficiently larger such that for all $n \ge N$, if $a_n \ge 0$ then $a_{n+1} \le 0$, and if $a_n \le 0$, then $a_{n+1} \ge 0$,
2. the terms eventually monotonically decrease in absolute value, i.e. there is some $N$ sufficiently large that for all $n \ge N$, we have $|a_{n+1}| \le |a_n|$, and
3. $\lim_{n\to\infty} a_n = 0$.

Then the series $$ \sum_{n=0}^{\infty} a_n $$ converges.

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Applying this theorem to your example, let $$ a_0 = 0, \qquad\text{and}\qquad a_n := (-1)^n\frac{n^{1/3}}{n} = (-1)^n n^{-2/3}. $$ Then

1. If $n \ge 1$, then $a_n \le 0$ whenever $n$ is odd, and $a_n \ge 0$ whenever $n$ is even; that is, the terms alternate signs, as required. This is not to difficult to check.
2. If $n \ge 1$, then $$ |a_{n+1}| = (n+1)^{-2/3} \le n^{-2/3} = |a_n|, $$ thus the terms decrease in absolute value, as required.
3. We have $$ \lim_{n\to \infty} a_n = \lim_{n\to \infty} (-1)^n n^{-2/3} = 0. $$

Therefore, by the Alternating Series Test, v2, the series $$ \sum_{n=0}^{\infty} a_n = \sum_{n=1}^{\infty} (-1)^{n} n^{-2/3} $$ converges.

Answer 2

The series is alternating with $a_n=n^{-\frac23}\to0$, hence convergent by the Leibniz test.

It can be put in the form you were taught by factoring out a $-1$. That is, $\sum_{n=1}^\infty (-1)^na_n=-\sum_{n=1}^\infty (-1)^{n+1}a_n$.

Answer 3

If you want to write it exactly as $\sum_{n=1}^{\infty} (-1)^{n+1} a_n$, then you can write like $-1 + \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{(n+1)^{\frac{2}{3}}}$