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The nth term of a series is represented by an=2^n/5^n+1 ⋅n . George correctly applies the ratio test to determine whether the series converges or diverges. Which statement reflects George's conclusion?


From the ratio test, r = 0.4. The series diverges.


From the ratio test, r = 0.4. The series converges.


From the ratio test, r = 4. The series converges.


From the ratio test, r = 4. The series diverges.

The nth term of a series is represented by an2n5n1 n George correctly applies the ratio test to determine whether the series converges or diverges Which stateme class=

Respuesta :

Answer:  Choice B) r = 0.4; series converges

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Explanation:

Check out the attached image below to see the steps on how I computed r.

The value you should get is r = 0.4

Since r is less than 1, the series converges.

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Extra info:

If r > 1, then the series would diverge.

If r = 1, then the series may diverge, conditionally converge, or absolutely converge. Another test would be needed if you get r = 1.

Ver imagen jimthompson5910
Cetacea

Answer: From the ratio test, r = 0.4. The series converges.

The given term is: [tex]a_{n}=\frac{2^{n}}{5^{\left(n+1\right)}}\cdot n[/tex]

So the next term is = [tex]a_{n+1}=\frac{2^{\left(n+1\right)}}{5^{\left(n+2\right)}}\cdot\left(n+1\right)[/tex]

The ratio test is :

[tex]\left|\frac{a_{n+1}}{a_{n}}\right|=\left|\frac{\frac{2^{\left(n+1\right)}}{5^{\left(n+2\right)}}\cdot\left(n+1\right)}{\frac{2^{n}}{5^{\left(n+1\right)}}\cdot n}\right|\\\\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_{n}}\right|=lim_{n\to\infty}\left|\frac{2^{\left(n+1\right)}}{2^{n}}\cdot\frac{5^{\left(n+1\right)}}{5^{\left(n+2\right)}}\cdot\frac{\left(n+1\right)}{n}\right|[/tex]

[tex]lim_{n\to\infty}\left|\frac{a_{n+1}}{a_{n}}\right|=lim_{n\to\infty}\left|\frac{2^{\left(n+1\right)}}{2^{n}}\cdot\frac{5^{\left(n+1\right)}}{5^{\left(n+2\right)}}\cdot\frac{\left(n+1\right)}{n}\right|[/tex]

[tex]lim_{n\to\infty}\left|\frac{a_{n+1}}{a_{n}}\right|=lim_{n\to\infty}\left|2\cdot\frac{1}{5}\cdot\frac{\left(n+1\right)}{n}\right|[/tex]

[tex]lim_{n\to\infty}\left|\frac{a_{n+1}}{a_{n}}\right|=\frac{2}{5}lim_{n\to\infty}\left|1+\frac{1}{n}\right|[/tex]

[tex]lim_{n\to\infty}\left|\frac{a_{n+1}}{a_{n}}\right|=\frac{2}{5}\left(1+0\right)[/tex]

[tex]lim_{n\to\infty}\left|\frac{a_{n+1}}{a_{n}}\right|=0.4[/tex]

Since 0.4 < 1 so the series converges.

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