Question

Define the sequences ${\left\{a_n\right\}}_{n=3}^{\infty }\text{and}{\left\{b_n\right\}}_{n=3}^{\infty }$ as

$a_n={\left(logn+loglogn\right)}^{logn}$ and $b_n=n^{\left(1+\frac{1}{logn}\right)}\mathrm{.}$

Which one of the following is TRUE?

1. ${\sum }_{n=3}^{\infty }\frac{1}{a_n}$ is convergent but ${\sum }_{n=3}^{\infty }\frac{1}{b_n}$ is divergent
2. ${\sum }_{n=3}^{\infty }\frac{1}{a_n}$ is divergent but ${\sum }_{n=3}^{\infty }\frac{1}{b_n}$ is convergent
3. Both ${\sum }_{n=3}^{\infty }\frac{1}{a_n}$ and ${\sum }_{n=3}^{\infty }\frac{1}{b_n}$ are divergent
4. Both ${\sum }_{n=3}^{\infty }\frac{1}{a_n}$ and ${\sum }_{n=3}^{\infty }\frac{1}{b_n}$ are convergent

Answer: ${\sum }_{n=3}^{\infty }\frac{1}{a_n}$ is convergent but ${\sum }_{n=3}^{\infty }\frac{1}{b_n}$ is divergent

Solution :-

Given that

We know that

Now, we can find a positive integer m such that

But we know that by p-series test ${\sum }_{n=3}^{\infty }\frac{1}{n^3}$ is convergent because p = 3 > 1.

So, by the comparison test ${\sum }_{n=3}^{\infty }\frac{1}{{\left(logn\right)}^{logn}}$ is also convergent.

Therefore, from eq(1), again by comparison test ${\sum }_{n=3}^{\infty }\frac{1}{a_n}={\sum }_{n=3}^{\infty }\frac{1}{{\left(log\left(n\cdot logn\right)\right)}^{logn}}$ is also convergent.

Now, we have $b_n=n^{\left(1+\frac{1}{logn}\right)}$

Here, we know that by p-series test ${\sum }_{n=3}^{\infty }\frac{1}{n}$ is divergent because p = 1 ≤ 1.

So, ${\sum }_{n=3}^{\infty }\frac{1}{b_n}$ is divergent.

So, the correct answer is option-1, $\to$ ${\sum }_{n=3}^{\infty }\frac{1}{a_n}$ is convergent but ${\sum }_{n=3}^{\infty }\frac{1}{b_n}$ is divergent