Chapter 8.6, Problem 36E

To determine

To find: The proof that the series ${\displaystyle \sum f_n\left(x\right)}$ converges for the value of $x$ but the series of ${\displaystyle \sum {f^{\prime }}_n\left(x\right)}$ diverges for $x=2n\pi$ $n$ is an integer, then determine the value of $x$ for which ${\displaystyle \sum {f^{″}}_n\left(x\right)}$ converges.

Answer to Problem 36E

Thefor the value of $x\ne k\pi$ ${\displaystyle \sum {f^{″}}_n\left(x\right)}$ diverges and proof that the series ${\displaystyle \sum f_n\left(x\right)}$ converges for the value of $x$ but the series of ${\displaystyle \sum {f^{\prime }}_n\left(x\right)}$ diverges for $x=2n\pi$ $n$ is an integer is obtained.

Explanation of Solution

Calculation:

Consider the series ${\displaystyle \sum f_n\left(x\right)}$ the nth term is $a_n=\frac{\sin x}{n^2}$ .

Consider the series $b_n$ where $b_n=\frac{1}{n^2}$ .

Here, $\left|a_n\right|\le \left|b_n\right|$ as $\sin nx\le 1$ .

Here, ${\displaystyle \sum b_n}$ is convergent p series as $p=2>1$ .

Then, by the comparison test the series ${\displaystyle \sum \frac{\sin x}{n^2}}$ and ${\displaystyle \sum f_n\left(x\right)}$ is convergent.

Consider the case for ${\displaystyle \sum {f^{\prime }}_n\left(x\right)}$ as,

  $\begin{array}{c}{f^{\prime }}_n=\frac{d}{dx}{f}_n\left(x\right)\\ =\frac{d}{dx}\frac{\sin nx}{n^2}\\ =\frac{\cos nx}{n^2}\\ =\frac{\cos n\left(2n\pi \right)}{n^2}\end{array}$

The above values converges to 1. Thus, the series is harmonic and divergent and for $x=2n\pi$ this proves that ${\displaystyle \sum {f^{\prime }}_n\left(x\right)}$ is divergent where n is the integer.

Consider the case for ${\displaystyle \sum {f^{″}}_n\left(x\right)}$ as,

  $\begin{array}{c}{f^{″}}_n\left(x\right)=\frac{d}{dx}{f^{\prime }}_n\left(x\right)\\ =\frac{d}{dx}\frac{\cos nx}{n}\\ =-\sin nx\end{array}$

In the above function for $x\ne k\pi$ , k is the integer, then $\underset{n\to \infty }{\lim }\sin nx$ does not tend to a unique finite values between $-1$ to 1 so for $x\ne k\pi$ . The series $-{\displaystyle \sum \sin nx}$ is divergent.