Question

question 6 of 13
determine convergence or divergence of $sum_{n = 1}^{infty}sin(\frac{1}{n})$ using any method.
the series converges.
the series diverges.

Explanation:

Step1: Use limit - comparison test

We know that for small \(x\), \(\sin x\sim x\) as \(x\to0\). Let \(a_n = \sin(\frac{1}{n})\) and \(b_n=\frac{1}{n}\). Then \(\lim_{n
ightarrow\infty}\frac{a_n}{b_n}=\lim_{n
ightarrow\infty}\frac{\sin(\frac{1}{n})}{\frac{1}{n}}\). Let \(x = \frac{1}{n}\), as \(n
ightarrow\infty\), \(x
ightarrow0\). And \(\lim_{x
ightarrow0}\frac{\sin x}{x}=1\).

Step2: Analyze the series of \(b_n\)

The series \(\sum_{n = 1}^{\infty}b_n=\sum_{n=1}^{\infty}\frac{1}{n}\) is the harmonic series, which is a well - known divergent series (\(p\) - series with \(p = 1\), and for \(p\) - series \(\sum_{n=1}^{\infty}\frac{1}{n^p}\), it diverges when \(p\leqslant1\)).

Step3: Apply the limit - comparison test result

Since \(\lim_{n
ightarrow\infty}\frac{\sin(\frac{1}{n})}{\frac{1}{n}} = 1>0\) and \(\sum_{n = 1}^{\infty}\frac{1}{n}\) diverges, by the limit - comparison test, the series \(\sum_{n=1}^{\infty}\sin(\frac{1}{n})\) diverges.

Answer:

The series diverges.