Question

How do you test the convergence of the series $\cos (n)\sin {\left( {\dfrac{\pi }{n}} \right)^2}$ ?

Answer 1

Hint: In this question, we need to test the convergence of the given series. We examine the convergence by direct comparison test. We make use of the fact that $\left| {\cos x} \right| \leqslant 1$ and $\left| {\sin x} \right| \leqslant \left| x \right|$.
Then we take the modulus of the given series and solve it further. Then we also make use of a p-series test to check the convergence of the one more series that we obtain. At last we apply the direct comparison test and verify whether it diverges or converges.

Complete step-by-step answer:
Given the series of the form $\cos (n)\sin {\left( {\dfrac{\pi }{n}} \right)^2}$
We are asked to examine the convergence of the series given above.
We check this by direct comparison test.
The direct comparison test states that if ${a_n}$ and ${b_n}$ are two series such that $0 \leqslant {a_n} \leqslant {b_n}$ for all n greater than the some positive integer N, then the following is true.
If $\sum\limits_{n = 1}^\infty {{b_n}} $ converges, then $\sum\limits_{n = 1}^\infty {{a_n}} $ also converges.
If $\sum\limits_{n = 1}^\infty {{b_n}} $ diverges, then $\sum\limits_{n = 1}^\infty {{a_n}} $ also diverges.
We have the inequalities related to the function sine and cosine.
We have that $\left| {\cos x} \right| \leqslant 1$ and $\left| {\sin x} \right| \leqslant \left| x \right|$.
Let us take modulus to the given series, we have, $\left| {\cos (n)\sin {{\left( {\dfrac{\pi }{n}} \right)}^2}} \right|$
Now by applying the inequalities we have,
$ \Rightarrow \left| {\cos (n)\sin {{\left( {\dfrac{\pi }{n}} \right)}^2}} \right| \leqslant \dfrac{{{\pi ^2}}}{{{n^2}}}$
Now we check the convergence of the series $\dfrac{{{\pi ^2}}}{{{n^2}}}$.
$\sum\limits_{n = 0}^\infty {\dfrac{{{\pi ^2}}}{{{n^2}}}} = {\pi ^2}\sum\limits_{n = 0}^\infty {\dfrac{1}{{{n^2}}}} $
Since ${\pi ^2}$ is constant, we have taken it out from the summation.
Now we apply the p-series test to the series $\sum\limits_{n = 0}^\infty {\dfrac{1}{{{n^2}}}} $.
We have the p-series test which states that the p-series $\sum\limits_{n = 1}^\infty {\left( {\dfrac{1}{{{n^p}}}} \right)} $ where $p > 0$ is convergent if and only if $p > 1$. Otherwise it diverges.
In this case we have $p = 2 > 1$, hence $\sum\limits_{n = 0}^\infty {\dfrac{1}{{{n^2}}}} $ is convergent.
Therefore we have, the series $\sum\limits_{n = 0}^\infty {\dfrac{{{\pi ^2}}}{{{n^2}}}} = {\pi ^2}\sum\limits_{n = 0}^\infty {\dfrac{1}{{{n^2}}}} $ is also convergent.
Now since $\sum\limits_{n = 0}^\infty {\dfrac{{{\pi ^2}}}{{{n^2}}}} $ is convergent, then by direct comparison test the series $\sum\limits_{n = 0}^\infty {\cos (n)\sin {{\left( {\dfrac{\pi }{n}} \right)}^2}} $ is also convergent and hence it is absolutely convergent.
Hence, the series $\cos (n)\sin {\left( {\dfrac{\pi }{n}} \right)^2}$ converges absolutely.

Note:
Students must know the meaning of convergence and divergence of a series.
The nth partial sum of the series $\sum\limits_{n = 1}^\infty {{a_n}} $ is given by ${S_n} = {a_1} + {a_2} + .... + {a_n}$. If the sequence of these partial sums $\{ {S_n}\} $ converges to L, then the sum of the series converges to L. If $\{ {S_n}\} $ diverges, then the sum of the series diverges.
The direct comparison test states that if ${a_n}$ and ${b_n}$ are two series such that $0 \leqslant {a_n} \leqslant {b_n}$ for all n greater than the some positive integer N, then the following is true.
If $\sum\limits_{n = 1}^\infty {{b_n}} $ converges, then $\sum\limits_{n = 1}^\infty {{a_n}} $ also converges.
If $\sum\limits_{n = 1}^\infty {{b_n}} $ diverges, then $\sum\limits_{n = 1}^\infty {{a_n}} $ also diverges.
The p-series test which states that the p-series $\sum\limits_{n = 1}^\infty {\left( {\dfrac{1}{{{n^p}}}} \right)} $ where $p > 0$ is convergent if and only if $p > 1$.