Question

sum_{n=2}^{infty} frac{1}{n} x^n 9. sum_{n=1}^{infty} frac{x^n}{n 3^n} 3, [-3, 3] 10. sum_{n=1}^{infty} n ratio test |frac{x^n cdot x}{(n+1) 3^n cdot 3} cdot frac{n 3^n}{x^n}| |frac{xn}{3(n+1)}| |frac{x}{3}| lim_{n o infty} |frac{n}{n+1}| = 1 |frac{x}{3}| < 1 |x| < 3 radius = 3 -3 < x < 3 sum_{n=1}^{infty} frac{-3^n}{n 3^n} sum frac{-1^n cdot 3^n}{}

          sum_{n=2}^{infty} frac{1}{n} x^n
9. sum_{n=1}^{infty} frac{x^n}{n 3^n} 3, [-3, 3]
10. sum_{n=1}^{infty} n
ratio test
|frac{x^n cdot x}{(n+1) 3^n cdot 3} cdot frac{n 3^n}{x^n}|
|frac{xn}{3(n+1)}|
|frac{x}{3}| lim_{n 	o infty} |frac{n}{n+1}| = 1
|frac{x}{3}| < 1
|x| < 3 radius = 3
-3 < x < 3
sum_{n=1}^{infty} frac{-3^n}{n 3^n}
sum frac{-1^n cdot 3^n}{}

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$sumn=2^infty frac1n x^n 9. sumn=1^infty fracx^nn 3^n 3, [-3, 3] 10. sumn=1^infty n ratio test |fracx^n cdot x(n+1) 3^n cdot 3 cdot fracn 3^nx^n| |fracxn3(n+1)| |fracx3| limn o infty |fracnn+1| = 1 |fracx3| < 1 |x| < 3 radius = 3 -3 < x < 3 sumn=1^infty frac-3^nn 3^n sum frac-1^n cdot 3^n$

Added by Joseph H.

Calculus

James Stewart 4th Edition

Chapter 8

Instant Answer

Solved by Expert Christopher Stanley

08/02/2022

Step 1

This resembles the harmonic series 1/n, which is known to be divergent. Show more…

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Thanks for your feedback!

When I’m doing this problem to find radius of convergence and interval of convergence, I plug in -3 and I can simplify right away to get -1/n so harmonic series test 1/n is divergent but the way it is solved is using the alternating series test and that give 0 which is convergent. Why can’t we just use the harmonic series test which gives divergent

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