Problem

Use any test to determine whether the series is a…

02:45

Problem 33 Medium Difficulty

Use any test to determine whether the series is absolutely convergent, conditionally convergent, or divergent.

$ \displaystyle \sum_{n = 1}^{\infty} \frac {( - 9)^n}{n10^{n+1}} $

Name: use any test to determine whether the series is absolutely convergent conditionally convergent or 3
Uploaded: 2018-10-20
Duration: 2 min 56 s
Channel: Numerade
Description: Use any test to determine whether the series is absolutely convergent, conditionally convergent, or divergent. $ \displaystyle \sum_{n = 1}^{\infty} \frac {( - 9)^n}{n10^{n+1}} $

Answer

CONVERGES

Topics

Sequences

Series

Calculus: Early Transcendentals

Chapter 11

Infinite Sequences and Series

Section 6

Absolute Convergence and the Ratio and Root Tests

Discussion

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Video Transcript

let's use the ratio test to show that the Siri's converges So we'll be looking at the limit and goes to infinity a n plus one over a end where the end is given by this term over here. So let's first deal with the numerator and plus one. So here will have you noticed that because we're taking absolute values, we can go ahead and ignore the negative nine over here and then we'LL have ten. And then there you were place and with n plus one and then add the other one over here to give you a plus two and then we'll divide that by am absolute value of that. So then, as usual here we'LL take that denominator and blue and flip it over Multiply it And then now we should cancel out as much as we can. We could take off end of these nines and we're left over with the nine when we do this on top. Similarly, we could cancel off and plus one of the tents that will leave you with one on the bottom and then we'll still have our and and plus one left over but in the limit and over and plus one goes toe one. So we get nine over ten, which is less than one. So we conclude that the Siri's the given Siri's converges. Absolutely. And then here we already mentioned what tests the ratio says, so that's our answer.

J H.