Question

How do you test the series? Sum for \[n=1\] to infinity \[{{\sin }^{2}}\left( \dfrac{1}{n} \right)\] convergence by using limit comparison test with \[{{c}_{n}}=\left( \dfrac{1}{{{n}^{2}}} \right)\] ?

Answer 1

Hint: These problems of convergence test are very easy to solve once we understand the underlying concepts behind the sum in depth. Solving this sum requires some prior knowledge of differential and integral calculus as well as some tests of convergence and their various applications in different scenarios. In these type of problems, we consider two series, say \[\sum\limits_{n=1}^{\infty }{{{a}_{n}}}\] and \[\sum\limits_{n=1}^{\infty }{{{b}_{n}}}\] , then if \[\displaystyle \lim_{n\to \infty }\dfrac{{{a}_{n}}}{{{b}_{n}}}=k\] , then either both the series converge or both of them diverge. This is known as the limit comparison test. If one of the two series is convergent, then we can also say that the other series is also convergent.

Complete step-by-step answer:
Now we start off with the solution to the given problem by writing the two series that we need to consider for this given problem, they are,
\[\sum\limits_{n=1}^{\infty }{{{\sin }^{2}}\left( \dfrac{1}{n} \right)}\] and \[\sum\limits_{n=1}^{\infty }{\left( \dfrac{1}{{{n}^{2}}} \right)}\] . Now we need to find the value of \[\displaystyle \lim_{n\to \infty }\dfrac{{{a}_{n}}}{{{b}_{n}}}=k\] . We substitute the respective values at the limit to get,
\[\displaystyle \lim_{n\to \infty }\dfrac{{{\sin }^{2}}\left( \dfrac{1}{n} \right)}{\left( \dfrac{1}{{{n}^{2}}} \right)}\]
Now in this problem we assume, \[\begin{align}
  & x=\dfrac{1}{n} \\
 & \therefore x \to 0,n\to \infty \\
\end{align}\]
Thus we can rewrite the limit equation as,
\[\displaystyle \lim_{x \to 0}\dfrac{{{\sin }^{2}}\left( x \right)}{\left( {{x}^{2}} \right)}\] . The value of this limit yields to \[\displaystyle \lim_{x \to 0}\dfrac{{{\sin }^{2}}\left( x \right)}{\left( {{x}^{2}} \right)}=1\] . From series and values we know that the series \[\sum\limits_{n=1}^{\infty }{\left( \dfrac{1}{{{n}^{2}}} \right)}\] is convergent as it yields a value equal to \[\dfrac{{{\pi }^{2}}}{6}\] . Thus we can clearly say that the other series \[\sum\limits_{n=1}^{\infty }{{{\sin }^{2}}\left( \dfrac{1}{n} \right)}\] is also convergent in nature.

Note: From problems like these we need to be through with our concepts of calculus and series and be able to find out the limiting value of any function. We also need to be very careful with our choice of the assumed function which is used to prove the convergence or divergence of the given function. We also must know about the convergence or divergence of the assumed series, failing which solving the problem might be cumbersome.