### 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)