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Question 15 Test the series below for convergence using the Root Test. $\sum_{n=1}^{\infty} \left(\frac{3n}{6n+3}\right)^n$ The limit of the root test simplifies to $\lim_{n \to \infty} |f(n)|$ where f(n) = The limit is: (enter oo for infinity if needed) Based on this, the series Converges Diverges Question Help: Video Add Work Check Answer Question 16 Estimate the error in using the partial sum of 45 terms of the series $\sum_{k=1}^{\infty} \frac{3}{k^3}$. Error ? Add Work Check Answer

          Question 15
Test the series below for convergence using the Root Test.
$\sum_{n=1}^{\infty} \left(\frac{3n}{6n+3}\right)^n$
The limit of the root test simplifies to $\lim_{n \to \infty} |f(n)|$ where
f(n) = 
The limit is:
(enter oo for infinity if needed)
Based on this, the series
Converges
Diverges
Question Help: Video
Add Work
Check Answer
Question 16
Estimate the error in using the partial sum of 45 terms of the series $\sum_{k=1}^{\infty} \frac{3}{k^3}$.
Error ?
Add Work
Check Answer

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Question 15
Test the series below for convergence using the Root Test.
∑n=1^∞((3n)/(6n+3))^n
The limit of the root test simplifies to limn →∞ |f(n)| where
f(n) =
The limit is:
(enter oo for infinity if needed)
Based on this, the series
Converges
Diverges
Question Help: Video
Add Work
Check Answer
Question 16
Estimate the error in using the partial sum of 45 terms of the series ∑k=1^∞(3)/(k^3).
Error ?
Add Work
Check Answer

Added by Jose N.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Drew Scalzo

Step 1

The Root Test states that if the limit of the absolute value of the nth root of the terms of the series is less than 1, then the series converges. In this case, we have the series x ≥ 31/(6n+3). To apply the Root Test, we need to find the limit of the absolute Show more…

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Can someone explain how to do this, explaining core concepts? Question 15 0/1 pt O 3 O Details Test the series below for convergence using the Root Test. x â‰¥ 31/(6n+3) The limit of the root test simplifies to lim(nâ†’âˆž) |(31/(6n+3))^(1/n)| where f(n) = 31/(6n+3). The limit is: {enter âˆž for infinity if needed} Based on this, the series O Converges O Diverges Question Help: Video Add Work Check Answer Question 16 0/1 pt O 3 O Details Estimate the error in using the partial sum of 45 terms of the series. Error Add Work Check Answer

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