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# Confirm that the Integral Test can be applied to the series. Then use the Integral Test to determine the convergence or divergence of the series.frac13+frac15+frac17+frac19+frac{1}{11}+...

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Series
asked 2021-02-09

Confirm that the Integral Test can be applied to the series. Then use the Integral Test to determine the convergence or divergence of the series.
$$\frac13+\frac15+\frac17+\frac19+\frac{1}{11}+...$$

## Answers (1)

2021-02-10
Given: The series
$$\frac13+\frac15+\frac17+\frac19+\frac{1}{11}+...$$
To determine: The convergence and divergence of the given series by Integral test.
Explanation:
Integral Test: Let f(x) be a positive, continuous and decreasing function for $$x\in[1,\infty)$$ and f(n)=a_n where n is any natural number then the improper integral $$\int_1^\infty f(x)dx$$ and infinite $$\sum_{n=1}^\infty a_n$$ both converge or diverge simultaneously.
Here the infinite sum is $$\frac13+\frac15+\frac17+\frac19+\frac{1}{11}+...=\sum_{n=1}^\infty\frac{1}{2n+1}$$
Therefore $$a_n=\frac{1}{2n+1}$$
So let $$f(x)=\frac{1}{2x+1}$$ for $$x\in[1,\infty)$$
Then clearly f(x) is continuous, positive, and decreasing for $$x\in[1,\infty)$$ and $$f(n)=\frac{1}{2n+1}=a_n$$ for any natural number n.
Hence Integral Test can be applied for $$f(x)=\frac{1}{2x+1}$$ and $$a_n\frac{1}{2n+1}$$
Now checking for the convergence or divergence of the improper integral
$$\int_1^\infty f(x)dx$$ for $$f(x)=\frac{1}{2x+1}$$
So,
$$\int_1^\infty f(x)dx$$
$$=\lim_{t\rightarrow\infty}\int_1^t\frac{1}{2x+1}dx$$
$$=\frac12\lim_{t\rightarrow\infty}\int_1^t\frac{2}{2x+1}dx$$ [multiplying and dividing by 2]
$$=\frac12\lim_{t\rightarrow\infty}[\ln(2x+1)]_1^t\quad[\because\int\frac{f'(x)}{f(x)}dx=\ln[f(x)]]$$
$$=\frac12\lim_{t\rightarrow\infty}[\ln(2t+1)-\ln3]$$
$$=\infty$$
Hence $$\int_1^\infty f(x)dx=\int_1^\infty\frac{1}{2x+1}dx$$ diverges
Therefore by Integral test
$$\sum_{n=1}^\infty\frac{1}{2n+1}=\frac13+\frac15+\frac17+\frac19+\frac{1}{11}+...$$ also diverge.
Answer: The series $$\frac13+\frac15+\frac17+\frac19+\frac{1}{11}+...$$ is divergent

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