Question

How do you determine whether the series is convergent or divergent given $ \sum{\dfrac{{{\sin }^{2}}\left( n \right)}{n\cdot \sqrt{n}}} $ for $ n=1 $ to $ n=\infty $ ?

Answer 1

Hint: We will use the fact that $ {{\sin }^{2}}\left( n \right)<1 $ to obtain a bound on the numerator. With this, we can come up with another infinite series that is greater than the given infinite series. We will check the convergence or divergence of the new series and use the direct comparison test to obtain the convergence or divergence of the given series.

Complete step by step answer:
The given series is $ \sum\limits_{n=1}^{\infty }{\dfrac{{{\sin }^{2}}\left( n \right)}{n\cdot \sqrt{n}}} $ . Let us consider the term $ {{\sin }^{2}}\left( n \right) $ . For any value of $ n $ , the value of $ {{\sin }^{2}}\left( n \right) $ is less than 1 for $ n=1 $ to $ n=\infty $ . That means, we have $ {{\sin }^{2}}\left( n \right)<1 $ . So, we can say that
 $ \dfrac{{{\sin }^{2}}\left( n \right)}{n\cdot \sqrt{n}}<\dfrac{1}{n\cdot \sqrt{n}} $
by comparing the fractions. Now, let us consider the following infinite series,
 $ \sum\limits_{n=1}^{\infty }{\dfrac{1}{n\cdot \sqrt{n}}} $
We can write $ n\cdot \sqrt{n}={{n}^{1}}\cdot {{n}^{\dfrac{1}{2}}} $ . We know the rule for exponents which states that $ {{a}^{m}}\cdot {{a}^{n}}={{a}^{m+n}} $ . Using this rule, we have $ n\cdot \sqrt{n}={{n}^{1+\dfrac{1}{2}}}={{n}^{\dfrac{3}{2}}} $ . We know that the infinite series $ \sum{\dfrac{1}{{{x}^{p}}}} $ converges if $ p>1 $ . Therefore, we can say that the series $ \sum\limits_{n=1}^{\infty }{\dfrac{1}{n\cdot \sqrt{n}}}=\sum\limits_{n=1}^{\infty }{\dfrac{1}{{{n}^{\dfrac{3}{2}}}}} $ converges since $ \dfrac{3}{2}>1 $ .
Next, we will look at the direct comparison test. The direct comparison test states the following,
(i) If the infinite series $ \sum{{{b}_{n}}} $ converges and $ 0\le {{a}_{n}}\le {{b}_{n}} $ for all sufficiently large $ n $ , then the infinite series $ \sum{{{a}_{n}}} $ also converges.
(ii) If the infinite series $ \sum{{{b}_{n}}} $ diverges and $ 0\le {{a}_{n}}\le {{b}_{n}} $ for all sufficiently large $ n $ , then the infinite series $ \sum{{{a}_{n}}} $ also diverges.
Since we have that $ \dfrac{{{\sin }^{2}}\left( n \right)}{n\cdot \sqrt{n}}<\dfrac{1}{n\cdot \sqrt{n}} $ and the series $ \sum\limits_{n=1}^{\infty }{\dfrac{1}{n\cdot \sqrt{n}}} $ converges, using the direct comparison test, we can say that the given series $ \sum\limits_{n=1}^{\infty }{\dfrac{{{\sin }^{2}}\left( n \right)}{n\cdot \sqrt{n}}} $ also converges.

Note:
It is important to understand the concept of convergence and divergence for series. There are different tests to check the convergence or divergence of a series. Some of these tests are the ratio test, root test, direct comparison test, limit comparison test, etc. We should be familiar with such tests as they are essential in determining the nature of a series.