Determining whether the series $\sum_{n=1}^{\infty} \dfrac{2^n+3}{n^{2}+1}$ converges or not

Question

Using either the Ratio Test or the Root Test, I want to know whether the following series converges or not $$\sum_{n=1}^{\infty} \dfrac{2^n+3}{n^{2}+1}$$

I don't think it would be possible to use the Root Test here, but perhaps the Ratio Test, however, I don't know how to simplify this when using $\dfrac{a_{n+1}}{a_{n}}$, it does not get simplified, if anything it makes things more complicated... can someone please help me figure this one out? I am not looking for a full solution, I just need a starting point, something that can help me use the Ratio or Root Test easily.. I am not sure if this can be solved using a different method, but these were the ones the book asked for.

Answer 1

Time to employ my favourite "trick"; which I refer to as Domination Leads To Irrelevancy.

In essence, when taking the limit of the quotient of two functions consisting of terms with different orders of convergence/divergence (constant, linear, quadratic, exponential, tetrational, etc), only the terms with the greatest orders matter. In this case; $$\lim_{n\to\infty}\frac{2^n+3}{n^2+1}=\lim_{n\to\infty}\frac{2^n}{n^2}$$

In this summation sense, it also means that $$\sum_{n\in\Bbb N}\frac{2^n+3}{n^2+1}\text{ and }\sum_{n\in\Bbb N}\frac{2^n}{n^2}$$ either both converge or both diverge.

You can evaluate which on the RHS here using the ratio test.

Answer 2

Hint: Use the fact that$$\sqrt[n]{\frac{2^n+3}{n^2+1}}=2\frac{\sqrt[n]{1+3\times2^{-n}}}{\sqrt[n]{n^2+1}}.$$

Answer 3

Essentially, you're looking at the sum of $2^n/n^2$. In fact,

$$\dfrac{2^n+3}{n^2+1} \ge \dfrac{2^n}{n^2+n^2} = \dfrac{1}{2}\dfrac{2^n}{n^2}$$

Since $2^n > n^2$ for $n>4$, the terms $a_n$ do not approach zero (in fact, they approach infinity!).