Question

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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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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