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Dirichlet test states that if

a_n is a decreasing sequence with $\lim_{n\to \infty}a_n=0$ \lim_{n\to \infty}a_n=0, and

b_n is a sequence such that its partial sums $B_m=\sum_{n=1}^m b_n$ B_m=\sum_{n=1}^m b_n is bounded, then the series $\sum_{n=1}^{\infty} a_nb_n$ \sum_{n=1}^{\infty} a_nb_n is converging.

We are going to test the convergence of $\sum_{n=1}^{\infty}\frac{\sin n}{n}$ \sum_{n=1}^{\infty}\frac{\sin n}{n}. Which of the following claims are True regarding the correct choices and statements about

a_n and

b_n for the Dirichlet Test?

With $a_n=\sin n$ a_n=\sin n, a_n is decreasing and $\lim_{n\to \infty}a_n=0$ \lim_{n\to \infty}a_n=0.

With $b_n=\sin n$ b_n=\sin n the partial sums $B_m=\sum_{n=1}^m \sin n$ B_m=\sum_{n=1}^m \sin n is bounded since

m is finite.

With $b_n=\sin n$ b_n=\sin n the partial sums

$B_m=\sum_{n=1}^m \sin n=\sum_{n=1}^m \frac{\cos(n-1)-\cos(n+1)}{2\sin 1}\\=\frac{1}{2 \sin 1}(\sum_{n=0}^{m-1} \cos n-\sum_{n=2}^{m+1} \cos n)\\=\frac{1}{2 \sin 1}(\cos 0+\cos 1-\cos m-\cos (m+1))$ B_m=\sum_{n=1}^m \sin n=\sum_{n=1}^m \frac{\cos(n-1)-\cos(n+1)}{2\sin 1}\\=\frac{1}{2 \sin 1}(\sum_{n=0}^{m-1} \cos n-\sum_{n=2}^{m+1} \cos n)\\=\frac{1}{2 \sin 1}(\cos 0+\cos 1-\cos m-\cos (m+1))

is bounded.

With $a_n=\frac{1}{n}$ a_n=\frac{1}{n}, a_n is decreasing and $\lim_{n\to \infty}a_n=0$ \lim_{n\to \infty}a_n=0.

With $b_n=\sin n$ b_n=\sin n the partial sums $|B_m|=|\sum_{n=1}^m \sin n|\leq \sum_{n=1}^m |\sin n|\leq \sum_{n=1}^m 1=m$ |B_m|=|\sum_{n=1}^m \sin n|\leq \sum_{n=1}^m |\sin n|\leq \sum_{n=1}^m 1=m is bounded.

With $b_n=\frac{1}{n}$ b_n=\frac{1}{n} the partial sums $B_m=\sum_{n=1}^m \frac{1}{n}$ B_m=\sum_{n=1}^m \frac{1}{n} is bounded since m is finite.

With $b_n=\frac{1}{n}$ b_n=\frac{1}{n} the partial sums $B_m=\sum_{n=1}^m \frac{1}{n}\leq \sum_{n=1}^m 1=m$ B_m=\sum_{n=1}^m \frac{1}{n}\leq \sum_{n=1}^m 1=m is bounded.

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In24-S1-MA1014 - Mathematics

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