1. \(\ds \infser \frac{2^n}{n!}\text{:}\) \begin{align*} \inflim{\frac{a_{n+1}}{a_n}} \amp = \lim_{n\to\infty}\frac{2^{n+1}/(n+1)!}{2^n/n!}\\ \amp = \lim_{n\to\infty} \frac{2^{n+1}n!}{2^n(n+1)!}\\ \amp = \lim_{n\to\infty} \frac{2}{n+1}\\ \amp =0. \end{align*} Since the limit is \(0\lt 1\text{,}\) by the Ratio Test \(\ds\infser \frac{2^n}{n!}\) converges. The fact that \(\inflim{\frac{a_{n+1}}{a_n}}=0\) can be interpreted to mean that in the long run, the term \(a_{n+1}\) is roughly \(0\) times as large as \(a_n\text{.}\) In other words, not only is \(a_n\) decreasing to \(0\text{,}\) it is decreasing very quickly. That is, the terms of \(a_n\) decrease to \(0\) sufficiently fast enough to guarantee the convergence of \(\infser a_n\text{.}\)

2. \(\ds\infser \frac{3^n}{n^3}\text{:}\) \begin{align*} \inflim{\frac{a_{n+1}}{a_n}} \amp = \lim_{n\to\infty} \frac{3^{n+1}/(n+1)^3}{3^n/n^3}\\ \amp = \lim_{n\to\infty}\frac{3^{n+1}n^3}{3^n(n+1)^3}\\ \amp = \lim_{n\to\infty} \frac{3n^3}{(n+1)^3}\\ \amp = 3. \end{align*} Since the limit is \(3>1\text{,}\) by the Ratio Test \(\ds\infser \frac{3^n}{n^3}\) diverges. The fact that \(\inflim{\frac{a_{n+1}}{a_n}}=3\) can be interpreted to mean that in the long run, the term \(a_{n+1}\) is roughly \(3\) times as large as \(a_n\text{,}\) so \(a_n\) is increasing by roughly a factor of \(3\) in the long run. We could also use Part 2 of Theorem 8.2.20 to determine that this series converges. The exponential will dominate the polynomial in the long runs, so \(\inflim 3^n/n^3=\infty\text{.}\)

3. \(\infser\frac{1}{n^2+1}\text{:}\) \begin{align*} \inflim\frac{a_{n+1}}{a_n} \amp = \lim_{n\to\infty} \frac{1/\big((n+1)^2+1\big)}{1/(n^2+1)}\\ \amp = \lim_{n\to\infty} \frac{n^2+1}{(n+1)^2+1}\\ \amp = 1. \end{align*} Since the limit is 1, the Ratio Test is inconclusive. We can easily show this series converges using the Integral Test. We can also use Direct Comparison Test or Limit Comparison Test, with each comparing to the series \(\ds \infser \frac{1}{n^2}\text{.}\)

Before we begin, be sure to note the difference between \((2n)!\) and \(2n!\text{.}\) When \(n=4\text{,}\) the former is \(8!=8\cdot7\cdot\ldots\cdot 2\cdot1=40,320\text{,}\) whereas the latter is \(2(4\cdot3\cdot2\cdot1) = 48\text{.}\)

Applying the Ratio Test: \begin{align*} \inflim\frac{a_{n+1}}{a_n} \amp = \lim_{n\to\infty} \frac{(n+1)!(n+1)!/\big(2(n+1)\big)!}{n!n!/(2n)!}\\ \amp = \lim_{n\to\infty}\frac{(n+1)!(n+1)!(2n)!}{n!n!(2n+2)!}\\ \end{align*} Noting that \((n+1)!=(n+1)\cdot n!\) and \((2n+2)! = (2n+2)\cdot(2n+1)\cdot(2n)!\text{,}\) we have \begin{align*} \amp = \lim_{n\to\infty}\frac{(n+1)(n+1)}{(2n+2)(2n+1)}\\ \amp = 1/4. \end{align*}

Since the limit is \(1/4\lt 1\text{,}\) by the Ratio Test we conclude \(\ds \infser \frac{n!n!}{(2n)!}\) converges.

To find the limit in the second to last line, recall that we just need to examine the leading terms of the numerator and denominator, which are \(n^2\) and \(4n^2\) respectively.

1. \begin{align*} \inflim \left(a_n\right)^{1/n} \amp = \inflim \left(\left(\frac{3n+1}{5n-2}\right)^n\right)^{1/n} \\ \amp =\inflim \frac{3n+1}{5n-2} = \frac 35\text{.} \end{align*} Since the limit is less than 1, we conclude the series converges. Note: it is difficult to apply the Ratio Test to this series.

2. \begin{align*} \inflim \left(a_n \right)^{1/n} \amp =\inflim \left(\frac{n^4}{(\ln(n))^n}\right)^{1/n} \\ \amp = \inflim \frac{\big(n^{4/n}\big)}{\ln(n)}\\ \end{align*} The limit of the numerator must be found using L'Hôpital's Rule for indeterminate powers \begin{align*} \inflim \left(n^{4/n}\right) \amp = \inflim e^{\ln\left(n^{4/n}\right)}\\ \amp = \inflim e^{{4\ln\left(n\right)}/n}\\ \end{align*} Now apply L'Hôpital's to the expression in the exponent: \begin{align*} \amp \stackrel{\ \text{ by LHR } \ }{=} \inflim e^{4/n}\\ \amp = e^0=1\text{.} \end{align*} Since the numerator approaches 1 (by L'Hôpital's Rule) and the denominator grows to infinity, we have \begin{equation*} \inflim \frac{\big(n^{4/n}\big)}{\ln(n)} =0\text{.} \end{equation*} Since the limit is less than 1, we conclude the series converges.

3. \(\inflim \left(\frac{2^n}{n^2}\right)^{1/n} = \inflim \frac{2}{\big(n^{2/n}\big)} = 2\text{.}\) Since this is greater than 1, we conclude the series diverges. Note: The Ratio Test is easy to apply to this series.

   (Also note: The limit in the denominator is found in a similar fashion as was illustrated in Part 2. In general \(\inflim (n)^{b/n}=1\) for any real number \(b\text{.}\))