Question

The following series do not satisfy the hypotheses of the alternating series test as stated. In each case, state which hypothesis is not satisfied. State whether the series converges absolutely.$\sum _{n=1}^{\mathrm{\infty }}(-1{)}^{n+1}\frac{{\sin }^{2}n}{n}$

Short answer

The series does not satisfy the monotonicity condition and does not converge absolutely.

Step-by-step solution

Identify Alternating Series Test Conditions

The alternating series test (Leibniz test) states that the series $\sum _{n=1}^{\mathrm{\infty }}(-1{)}^{n+1}{a}_n$ converges if: 1) $a_n>0$ for all $n$, 2) $a_n\to 0$ as $n\to \mathrm{\infty }$, and 3) $a_{n+1}\le a_n$.

Examine Series Terms

For the given series $\sum _{n=1}^{\mathrm{\infty }}(-1{)}^{n+1}\frac{{\sin }^{2}n}{n}$, the terms $a_n=\frac{{\sin }^{2}n}{n}$ are functions of $n$. Note that $0\le {\sin }^{2}n\le 1$, so the terms $a_n$ are positive.

Check Limit Condition

We need $\underset{n\to \mathrm{\infty }}{lim}\frac{{\sin }^{2}n}{n}=0$. As $n\to \mathrm{\infty }$, since ${\sin }^{2}n$ is bounded between 0 and 1, and $n\to \mathrm{\infty }$, clearly $\frac{{\sin }^{2}n}{n}\to 0$. This condition is satisfied.

Verify Monotonicity

We must check if $\frac{{\sin }^2(n+1)}{n+1}\le \frac{{\sin }^{2}n}{n}$. However, since ${\sin }^{2}n$ can oscillate between 0 and 1 for different $n$, this inequality isn't necessarily true for every $n$. Hence, this monotonicity condition is not satisfied.

Determine Absolute Convergence

For absolute convergence, consider $\sum _{n=1}^{\mathrm{\infty }}|(-1{)}^{n+1}\frac{{\sin }^{2}n}{n}|=\sum _{n=1}^{\mathrm{\infty }}\frac{{\sin }^{2}n}{n}$. By the comparison test with the harmonic series $\sum \frac{1}{n}$ (which is divergent), since ${\sin }^{2}n\le 1$, the absolute convergence test fails.

Key concepts

Convergence

Convergence in the context of an alternating series refers to whether the series approaches a finite limit as more terms are added. For the alternating series test, convergence is assured if the series satisfies a few key conditions. Let's break these down:

• The terms you are adding, represented as $a_n$, must be positive. In simpler words, each term should be greater than zero if you omit the alternating sign factor $(-1{)}^{n+1}$.


• It must be shown that $a_n$ approaches zero as $n$ goes to infinity. This essentially means that each term becomes negligibly small, ensuring that their sum stabilizes rather than grows indefinitely.


• The terms in the series should be non-increasing, i.e., each term should be smaller than or equal to the one before it. This condition prevents any sudden large increases in term size, which could disrupt convergence.

In our specific series, we see that the first two conditions are met, but the third is not, due to the oscillating nature of ${\sin }^{2}n$. This affects its monotonicity which directly influences convergence.

Monotonicity

Monotonicity concerns the behavior of a sequence in relation to its order. For a series to pass the alternating series test, its terms $a_n$ must be non-increasing. This means $a_{n+1}\le a_n$ for every term in the sequence.

For the series $\sum _{n=1}^{\mathrm{\infty }}(-1{)}^{n+1}\frac{{\sin }^{2}n}{n}$, we see that ${\sin }^{2}n$ can fluctuate unpredictably between 0 and 1 as $n$ changes. This causes the numerator of the terms $\frac{{\sin }^{2}n}{n}$ to potentially increase, violating monotonicity. Thus, not every term is guaranteed to be smaller than the preceding one.

It is the unpredictable variation of ${\sin }^{2}n$ that breaks this monotonic requirement, demonstrating why the alternating series test might not establish convergence despite the alternating nature.

Absolute Convergence

Absolute convergence is a stricter form of convergence where the series composed of the absolute values of its terms also converges. To check for absolute convergence, consider the series $\sum _{n=1}^{\mathrm{\infty }}|(-1{)}^{n+1}\frac{{\sin }^{2}n}{n}|=\sum _{n=1}^{\mathrm{\infty }}\frac{{\sin }^{2}n}{n}$.

For absolute convergence, one often uses comparison tests. In this scenario, we compare our series with the harmonic series, $\sum \frac{1}{n}$, a known divergent series. Since ${\sin }^{2}n\le 1$, it means that $\frac{{\sin }^{2}n}{n}\le \frac{1}{n}$. However, because the harmonic series diverges and $\sum \frac{{\sin }^{2}n}{n}$ cannot be shown to be smaller than a convergent series, our series does not converge absolutely.

Without absolute convergence, it further underscores the limits on convergence claimed by the alternating series test, reaffirming why absolute convergence is a useful and powerful criterion in assessing series.