There are many different tests that can be used to test whether a given series converges or diverges. As a result, it is important to develop a good sense for quickly narrowing down which tests make sense to apply to a given series just by inspecting the given series. A fundamentally important fact allows us to gain some insight into the Ratio Test:
To gain a little practice, suppose that \(p(x) = x^5\). Then the limit to evaluate is:
\[ \lim _{n \to \infty } \frac {p(n+1)}{p(n)} = \answer {\frac {(n+1)^5}{n^5}} \]
The numerator is a polynomial of degree \(\answer {5}\), and the coefficient of the \(n^5\) term is \(\answer {1}\).

The denominator is a polynomial of degree \(\answer {5}\), and the coefficient of the \(n^5\) term is \(\answer {1}\).

Hence, the limit is \(\answer {1}\).

To explore this a little more, now set \(p(x) = x^5+4x^3+7\). Then:
\[ \lim _{n \to \infty } \frac {p(n+1)}{p(n)} = \answer {\frac {(n+1)^5+4(n+1)^3+7}{n^5+4n^3+7}} \]

Note that \((n+1)^5\) is a polynomial of degree \(\answer {5}\), \((n+1)^3\) is a polynomial of degree \(\answer {3}\), so without expanding any of these powers out, the numerator is a polynomial of degree \(\answer {5}\), and the coefficient of the \(n^5\) term is \(\answer {1}\).

The denominator is a polynomial of degree \(\answer {5}\), and the coefficient of the \(n^5\) term is \(\answer {1}\).

Hence, the limit is \(\answer {1}\).

One of the major consequences of this is that multiplying or dividing a given sequence by a polynomial will not affect the limit required in the Ratio Test. Indeed, consider the series \(\sum _{k=1}^{\infty } \frac {1}{2^k}\). Note that this is a geometric series, but we could establish its convergence using the Ratio Test. Indeed, here:
\[ L = \lim _{n \to \infty } \frac {a_{n+1}}{a_n} = \lim _{n \to \infty } \frac {1}{2^{n+1}}\cdot 2^n = \answer {\frac {1}{2}} \]

Now, let’s multiply the original sequence defined by the rule \(a_n = \left (\frac {1}{2}\right )^n\) by the polynomial \(p(n) = n^5\) and try to sum its terms. That is, we want to consider the series \(\sum _{k=1}^{\infty } \frac {k^5}{2^k}\). We find:

\[ L = \lim _{n \to \infty } \frac {(n+1)^5}{2^{n+1}} \cdot \frac {2^n}{n^5} = \lim _{n \to \infty } \frac {2^n}{2^{n+1}} \cdot \frac {(n+1)^5}{n^5} = \answer {\frac {1}{2}} \cdot \answer {1} \]

Introducing the \(n^5\) term into the numerator did not affect the value of \(L\).

Indeed, for the series \(\sum _{k=1}^{\infty } \frac {1}{2^k}\), we found that \(L=\answer {\frac {1}{2}}\) and for the series \(\sum _{k=1}^{\infty } \frac {k^5}{2^k}\), we found \(L=\answer {\frac {1}{2}}\).

One of the major consequences of this now is that a series must have a term that grows at least exponentially in order for the Ratio Test to have a chance to be conclusive! More explicitly, let \(p, q > 0\), \(a>1\), and consider the growth rates results for sequences:

Without performing any calculations, determine which of the following series would the Ratio Test would be conclusive. That is, which of the following would either converge or diverge as a consequence of the Ratio Test?

\(\sum _{k=1}^{\infty } \frac {k^3+2k-1}{k^2+2}\) \(\sum _{k=1}^{\infty } \frac {\ln (k) +k^2}{k!}\) \(\sum _{k=1}^{\infty } \frac {k^{10}}{5^k}\) \(\sum _{k=1}^{\infty } \frac {(\ln k)^{80}}{k^5}\) \(\sum _{k=1}^{\infty } k^{2018}\) \(\sum _{k=1}^{\infty } \frac {2^k}{k^k}\)