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Problem 3: Determine if the series converges absolutely, conditionally, or diverges. a. $\sum_{n=1}^{\infty} \frac{(-1)^n 4n}{\sqrt{n^3+3}}$ b. $\sum_{n=1}^{\infty} \frac{n^2}{3n^2 - 6}$ c. $\sum_{m=1}^{\infty} (-1)^m [\ln(2m+1) - \ln(m+3)]$ d. $\sum_{n=1}^{\infty} \frac{(-1)^n \sqrt{n}}{2n+3}$ e. $\sum_{n=1}^{\infty} (-1)^n \frac{\sqrt{2}}{n^{1/n}}$ f. $\sum_{n=1}^{\infty} \tan^{-1}((-1)^n)$

          Problem 3: Determine if the series converges absolutely, conditionally, or diverges.
a. $\sum_{n=1}^{\infty} \frac{(-1)^n 4n}{\sqrt{n^3+3}}$
b. $\sum_{n=1}^{\infty} \frac{n^2}{3n^2 - 6}$
c. $\sum_{m=1}^{\infty} (-1)^m [\ln(2m+1) - \ln(m+3)]$
d. $\sum_{n=1}^{\infty} \frac{(-1)^n \sqrt{n}}{2n+3}$
e. $\sum_{n=1}^{\infty} (-1)^n \frac{\sqrt{2}}{n^{1/n}}$
f. $\sum_{n=1}^{\infty} \tan^{-1}((-1)^n)$

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$problem 3 determine if the series converges absolutely conditionally or diverges a sumn1infty frac 1n 4nsqrtn33 b sumn1infty fracn23n2 6 c summ1infty 1m ln2m1 lnm3 d sumn1infty frac 1n s 51571$

Added by Madison W.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Sam Stansfield

12/15/2023

Step 1

a. $\sum_{n=1}^{\infty} \frac{(-1)^n 4n}{\sqrt{n^3+3}}$: We can use the alternating series test. Let $a_n = \frac{4n}{\sqrt{n^3+3}}$. Then $\lim_{n\to\infty} a_n = 0$ and $a_n$ is decreasing. Thus, the series converges conditionally. To check for absolute Show more…

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Problem 3: Determine if the series converges absolutely, conditionally, or diverges. a. $\sum_{n=1}^{\infty} \frac{(-1)^n 4n}{\sqrt{n^3+3}}$ b. $\sum_{n=1}^{\infty} \frac{n^2}{3n^2 - 6}$ c. $\sum_{m=1}^{\infty} (-1)^m [\ln(2m+1) - \ln(m+3)]$ d. $\sum_{n=1}^{\infty} \frac{(-1)^n \sqrt{n}}{2n+3}$ e. $\sum_{n=1}^{\infty} (-1)^n \frac{\sqrt{2}}{n^{1/n}}$ f. $\sum_{n=1}^{\infty} \tan^{-1}((-1)^n)$

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Transcript

00:01 Okay, we're looking at the convergence of various infinite series.

00:06 So this one, i'm going to use the comparison test.

00:15 Okay, and here's why.

00:17 The inverse tangent of n is always less than, it's actually always strictly less than pi over 2.

00:27 Because it's only equal to pi over 2 when the limit of n goes to infinity.

00:31 So any specific value of n is going to be less than that.

00:37 So that means that this is strictly less than 1 over n to the 3 halves power.

00:53 And this, that's a convergent p -series.

01:02 So therefore it converges.

01:05 Because it's less than a convergent p -series.

01:10 Okay, so for the next one, we have the exponential of sine of 1 over n.

01:14 So as n goes to infinity, 1 over n goes to 0.

01:19 So the sine of 1 over n goes to 0 as well.

01:26 So if we take this limit, that equals 1.

01:43 Because 1 over n goes to 0, so the sine of 1 over n goes to 0 as well, and e to the 0 is 1.

01:53 So that goes to 1, and so therefore it diverges because the limit of that sequence has to be 0 for it to converge.

02:08 Of course that's not a sufficient condition to prove convergence, but it doesn't converge.

02:17 Okay, for this one i'm going to use the ratio test.

02:22 So we have to take this limit, and we take the n plus first term divided by the nth term.

02:48 So resolve the complex fraction.

03:10 So if we take that limit, so n factorial over n plus 1 factorial.

03:17 So if i just do this one more time...

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