Question

Use the Limit Comparison Test to determine the convergence or divergence of the series. sum_{n=1}^inftysinfrac{1}{n}

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asked 2021-01-19
Use the Limit Comparison Test to determine the convergence or divergence of the series.
\(\sum_{n=1}^\infty\sin\frac{1}{n}\)

Expert Answers (1)

2021-01-20
Limit comparison test:
Suppose we have two series \(\sum a_n\) and \(\sum b_n\). Define
\(\lim_{n\to\infty}\frac{a_n}{b_n}=L\)
If L is positive and finite then either both series convergent or both series divergent.
Given that,
\(\sum_{n=1}^\infty\sin\frac{1}{n}\)
Since we have,
\(\sum_{n=1}^\infty\sin\frac{1}{n}\)
\(=\sum_{n=1}^\infty(\frac{1}{n}-(\frac{1}{n})^3\frac{1}{3!}+(\frac{1}{n})^5\frac{1}{5!}+...)\)
[Using \(\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}+...\)]
Here
\(a_n=(\frac{1}{n}-(\frac{1}{n})^3\frac{1}{3!}+(\frac{1}{n})^5\frac{1}{5!}+...)\)
So
\(b_n=\frac{1}{n}\)
Now evaluate \(\lim_{n\to\infty}\frac{a_n}{b_n}\), so
\(\lim_{n\to\infty}\frac{a_n}{b_n}=\lim_{n\to\infty}\frac{\frac{1}{n}-(\frac{1}{n})^3\frac{1}{3!}+(\frac{1}{n})^5\frac{1}{5!}+...}{\frac{1}{n}}\)
\(=\lim_{n\to\infty}\frac{\frac{1}{n}[1-(\frac{1}{n})^2\frac{1}{3!}+(\frac{1}{n})^4\frac{1}{5!}+...]}{\frac{1}{n}}\)
\(=\lim_{n\to\infty}[1-(\frac{1}{n})^2\frac{1}{3!}+(\frac{1}{n})^4\frac{1}{5!}+...]\)
\(=1\)
So \(\lim_{n\to\infty}\frac{a_n}{b_n}=1>0\) which is positive and finite.
Using p series test [\(\sum\frac{1}{n^p}\) is convergent if p>1 or divergent if \(p\leq1\)], \(\sum b_n\) is divergent.
Since \(\lim_{n\to\infty}\frac{a_n}{b_n}=1>0\) is positive and finite. \(\sum b_n\) is divergent, so using limit comparison test \(\sum a_n\) is also divergent.
Hence \(\sum_{n=1}^\infty\sin\frac{1}{n}\) is divergent.
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