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Real Analysis Question on Sequences and Series

For n ∈Nn\ \in \N, let
an=1(3n+2)(3n+4)a_n=\frac{1}{(3n+2)(3n+4)} and bn=n3+cos⁡(3n)3n+n3b_n=\frac{n^3+\cos(3^n)}{3n+n^3}.
Then, which one of the following is TRUE ?

A

∑n=1∞an\sum\limits_{n=1}^{\infin}a_n is convergent but ∑n=1∞bn\sum\limits_{n=1}^{\infin}b_n is divergent

B

∑n=1∞an\sum\limits_{n=1}^{\infin}a_n is divergent but ∑n=1∞bn\sum\limits_{n=1}^{\infin}b_n is convergent

C

Both ∑n=1∞an\sum\limits_{n=1}^{\infin}a_n and ∑n=1∞bn\sum\limits_{n=1}^{\infin}b_n are divergent

D

Both ∑n=1∞an\sum\limits_{n=1}^{\infin}a_n and ∑n=1∞bn\sum\limits_{n=1}^{\infin}b_n are convergent

Answer

Both ∑n=1∞an\sum\limits_{n=1}^{\infin}a_n and ∑n=1∞bn\sum\limits_{n=1}^{\infin}b_n are convergent

Explanation

Solution

The correct option is (D) : Both ∑n=1∞an\sum\limits_{n=1}^{\infin}a_n and ∑n=1∞bn\sum\limits_{n=1}^{\infin}b_n are convergent.