Question

For $n\in \mathbb{N}$ , let

$a_n=\frac{1}{\left(3n+2\right)\left(3n+4\right)}\text{and}{b}_n=\frac{n^3+cos\left(3^n\right)}{3^n+n^3}\cdot$

Then, which one of the following is TRUE?

1. $\sum _{n=1}^{\infty }{a}_n\text{is convergent but}\sum _{n=1}^{\infty }{b}_n\text{is divergent}$
2. $\sum _{n=1}^{\infty }{a}_n\text{is divergent but}\sum _{n=1}^{\infty }{b}_n\text{is convergent}$
3. $\text{Both}\sum _{n=1}^{\infty }{a}_n\text{and}\sum _{n=1}^{\infty }{b}_n\text{are divergent}$
4. $\text{Both}\sum _{n=1}^{\infty }{a}_n\text{and}\sum _{n=1}^{\infty }{b}_n\text{are convergent}$

Answer: $\text{Both}\sum _{n=1}^{\infty }{a}_n\text{and}\sum _{n=1}^{\infty }{b}_n\text{are convergent}$

Solution :-

Given that $a_n=\frac{1}{\left(3n+2\right)\left(3n+4\right)}$ , let $c_n=\frac{1}{n^2}$ .

So, By comparison test (limit form test), both $\sum _{n=1}^{\infty }{a}_n$ and $\sum _{n=1}^{\infty }{c}_n$ behave alike.
But $\sum _{n=1}^{\infty }{c}_n=\sum _{n=1}^{\infty }\frac{1}{n^2}$ is convergent by p-series test.

Hence, $\sum _{n=1}^{\infty }{a}_n$ is also convergent.

Now, Given that $b_n=\frac{n^3+cos\left(3^n\right)}{3^n+n^3}$ .

We know that

Now, let $k_n=\frac{n^3+1}{3^n}$ so, $k_{n+1}=\frac{{\left(n+1\right)}^3+1}{3^{n+1}}$ .

So, By D’Alembert’s Ratio test, $\sum _{n=1}^{\infty }{k}_n$ is convergent.

And hence, By comparison test $\sum _{n=1}^{\infty }{b}_n=\sum _{n=1}^{\infty }\frac{n^3+cos\left(3^n\right)}{3^n+n^3}$ is convergent.

So, the correct answer is option-(D), $\to \text{Both}\sum _{n=1}^{\infty }{a}_n\text{and}\sum _{n=1}^{\infty }{b}_n\text{are convergent}$