### Instructions:Select the FIRST correct reason why the given series converges or choose E for Diverges.**Options:**- **A.** Comparison with convergent geometric series- **B.** Convergent p series- **C.** Comparison with a convergent p series- **D.** Converges by limit comparison test- **E.** Diverges### Series:1. $\sum_{k=1}^{\infty} \frac{1}{(k+2)(k+3)}$2. $\sum_{k=1}^{\infty} \sin\left(\frac{1}{k}\right)$3. $\sum_{k=1}^{\infty} \frac{(k+1)^3}{k^{9/2}}$4. $\sum_{k=1}^{\infty} \frac{3k^2 + 2}{k^2 + 3k + 2}$5. $\sum_{k=2}^{\infty} \frac{1}{k^2 + 3k + 2}$6. $\sum_{k=1}^{\infty} \frac{k+1}{k^2 + 1}$7. $\sum_{k=2}^{\infty} \frac{1}{(k+3)(\ln(k))^{0.9}}$8. $\sum_{k=2}^{\infty} \frac{\ln(k)}{k^2}$9. $\sum_{k=2}^{\infty} \frac{1}{\sqrt{k \ln(k)}}$10. $\sum_{k=1}^{\infty} \frac{\ln(k)}{\sqrt{2k + 3}}$

NOTE: According to guideline answer of first three subpart can be given, for other please ask in a different question and specify the question number . 1)∑k=1∞ 1k+2k+3let,ak=1k+2k+3=1k2+5k+6bk=1k2Then, limk→∞akbk=limk→∞k2k2+5k+6=1As ∑k=1∞bk converges by P-seriesso,by limit comparison test ∑k=1∞ak is convergent∴(D)2) ∑k=1∞ sin 1kAs 2πx< sin(x) x∈0,π2⇒2π1k<sin 1k k∈N⇒2π∑k=1∞1k < ∑k=1∞ sin 1kAgain, ∑k=1∞1k diverges by P-series testso,∑k=1∞ sin 1k is divergent∴(E) 3) ∑k=1∞k+13k92Let, ak=k+13k92=k31+1k3k92=1+1k3k1.5bk=1k1.2Then, ak<bk k∈N &∑k=1∞ bk is convergent by P-seriesso,∑k=1∞ ak is convergent by comparison with a convergent P-series∴(C)

Math

Advanced Math

Select the FIRST correct reason why the given series converges or choose E for Diverges. A. Comparison with convergent geometric series B. Convergent p series C. Comparison with a convergent p series D. Converges by limit comparison test E. Diverges

Select the FIRST correct reason why the given series converges or choose E for Diverges. A. Comparison with convergent geometric series B. Convergent p series C. Comparison with a convergent p series D. Converges by limit comparison test E. Diverges

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

### Instructions:
Select the FIRST correct reason why the given series converges or choose E for Diverges.

**Options:**
- **A.** Comparison with convergent geometric series
- **B.** Convergent p series
- **C.** Comparison with a convergent p series
- **D.** Converges by limit comparison test
- **E.** Diverges

### Series:

1. $\sum_{k=1}^{\infty} \frac{1}{(k+2)(k+3)}$

2. $\sum_{k=1}^{\infty} \sin\left(\frac{1}{k}\right)$

3. $\sum_{k=1}^{\infty} \frac{(k+1)^3}{k^{9/2}}$

4. $\sum_{k=1}^{\infty} \frac{3k^2 + 2}{k^2 + 3k + 2}$

5. $\sum_{k=2}^{\infty} \frac{1}{k^2 + 3k + 2}$

6. $\sum_{k=1}^{\infty} \frac{k+1}{k^2 + 1}$

7. $\sum_{k=2}^{\infty} \frac{1}{(k+3)(\ln(k))^{0.9}}$

8. $\sum_{k=2}^{\infty} \frac{\ln(k)}{k^2}$

9. $\sum_{k=2}^{\infty} \frac{1}{\sqrt{k \ln(k)}}$

10. $\sum_{k=1}^{\infty} \frac{\ln(k)}{\sqrt{2k + 3}}$

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