0

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.

Friedrich
  • 487

3 Answers3

4

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!).

David P
  • 12,320
1

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

José Carlos Santos
  • 427,504
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.

Rhys Hughes
  • 12,842
  • This does not hold in general. $\lim (1/n^2) = \lim (1/n)$ – David P Jul 04 '20 at 07:28
  • The statement of the DLTI trick is not that $$\lim A=\lim B \implies \sum A, \sum B \text { both conv/div}$$ As you demonstrate, that is not true in general. However, that isn't seen here. What we see here is two polynomials $P(n), Q(n)$ with dominant term $p(n), q(n)$, and DLTI does apply in this case to say that $$\lim\frac PQ =\lim \frac pq$$ and also that $$\sum\frac PQ \text{ and } \sum\frac pq \text{ both conv/div}$$ – Rhys Hughes Jul 04 '20 at 13:24
  • The 'dominant term' is the function $p(n)$, rather than the value, toward which the function $P(n)$ tends as $n\to \infty$. In any polynomial, the dominant term is the term of the highest degree (all the rest become "irrelevant" insofar as taking limits), in the case of our $P(n)=2^n+3$, it is obvious that the exponential $2^n$ is the dominant term. – Rhys Hughes Jul 04 '20 at 13:37