# To test the series e 2n for convergence, you can use the Integral Test. (This is...

###### Question:

To test the series e 2n for convergence, you can use the Integral Test. (This is also a geometric series, so we could n=1 also investigate convergence using other methods.) Find the value of e-24 dx = Preview Ji What does this value tell you about the convergence of the series e-2n? the series definitely diverges the series might converge or diverge: we need more information the series definitely converges
Compute the value of the following improper integral, if it converges. (If it diverges, enter oo if it diverges to infinity, - oo if it diverges to negative infinity, or DNE if it diverges for some other reason.) Hint: integrate by parts. poo 3 ln(2) Preview 72025 po 3 ln(n) What does the value of the improper integral tell use about the convergence of the series 2 m2 %? O the series diverges O the series converges the Integral Test does not apply
Compute the value of the improper integral. (If the integral diverges to oo, type oo; if the integral diverges to - oo, type -00; and if the integral diverges for some other reason, type DNE.) pao da Preview J2 (4.2 + 4) 6 = Use your answer to help determine whether the series 1 converges or diverges. Enter C if the series is n=2 (4n + 4) convergent, D if the series is divergent, or ? if the Integral Test does not apply:
, type -oo; if it Compute the value of the improper integral. (If the integral diverges to oo, type oo; if it diverges to- diverges for some other reason, type DNE.) 3 Preview 2 dx = Ji 1 + 003 caries Use the value of the improper integral to determine whether or not the series 5 converges or diverges. Enter n=11+n2 Converges o C if the series is convergent, D if the series is divergent, and ? if the Integral Test does not apply:
To test the series for convergence, you can use the P-test. (You could also use the Integral Test, as is the case k=1 with all series of this type.) According to the P-test: diverges k=1 O the P-test does not apply to k= 1 obfeurge Now compute 84, the partial sum consisting of the first 4 terms of k= 1 84 = Preview
To test the series for convergence, you can use the P-test. (You could also use the Integral Test, as is the case k1 1.3 with all series of this type.) According to the P-test: E converges the P-test does not apply to Στε kvk I e diverges diverges Now compute 85, the partial sum consisting of the first 5 terms of IM S5 = Preview
Estimate the error in using the partial sum of 25 terms of the series Error < Preview
Find the smallest number n of terms needed to obtain an approximation of the series 27 accurate to 10 -4. n =
Find the smallest number n of terms needed to obtain an approximation of the series 26ke .lok accurate to 10-6 k=1 n =
To test this series for convergence * 3" +1 251 You could use the Limit Comparison Test, comparing it to the series ph where r= Preview n = 1 Completing the test, it shows the series: Converges Diverges
To test this series for convergence You could use the Limit Comparison Test, comparing it to the series where p= Preview np n= 1 Completing the test, it shows the series: Diverges Converges
To test the series 210 for convergence, you will use the Limit Comparison Test, comparing it to the p series 2k8 11 Στο where p= Preview k= 1 00 Now by the limit comparison test, the series ) k 2k8 + 51.10 converges diverges Licenco
We want to use the Basic Comparison Test (sometimes called the Direct Comparison Test or just the Comparison Test) to determine if the series: IM k5 - 3 converges or diverges by comparing it with: Σ We can conclude that: The Basic Comparison Test is inconclusive in this situation. The first series converges by comparison with the second series. The first series diverges by comparison with the second series.

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