Calculus Early Transcendentals: Integral & Multi-Variable Calculus for Social Sciences

Determine whether each series converges absolutely, converges conditionally, or diverges.

1. \(\ds\sum_{n=1}^\infty (-1)^{n-1}{1\over 2n^2+3n+5}\)

   Answer
   converges absolutely
   Solution
   We first notice that the series

   \begin{equation*} \sum_{n=1}^{\infty} \frac{1}{2n^2+3n+5} \end{equation*}

   converges by comparison to the convergent \(p\)-series \(\sum \frac{1}{n^2}\text{.}\) Next, we check that

   \begin{equation*} a_n = \frac{1}{2n^2+3n+5} > 0, \end{equation*}

   and

   \begin{equation*} \lim_{n\to\infty} a_n = 0, \end{equation*}

   and

   \begin{equation*} a_{n+1} \leq a_n \implies 2(n+1)^2 + 3(n+1) + 5 \geq 2n^2 + 3n +5, \end{equation*}

   which is true. Therefore, the series converges absolutely.

2. \(\ds\sum_{n=1}^\infty (-1)^{n-1}{3n^2+4\over 2n^2+3n+5}\)

   Answer
   diverges
   Solution
   Since

   \begin{equation*} \begin{split}\lim_{n\to\infty} a_n \amp = \lim_{n\to\infty} \frac{3n^2+4}{2n^2+3n+5} \\ \amp \Heq \lim_{n\to\infty} \frac{6n}{4n+3}\\ \amp \Heq \lim_{n\to\infty} \frac{6}{4} \neq 0,\end{split} \end{equation*}

   by the \(n\)-th Term Test, the series diverges.

3. \(\ds\sum_{n=1}^\infty (-1)^{n-1}{\ln n\over n}\)

   Answer
   converges conditionally
   Solution

   For \(n\geq 4\text{,}\) \(\ln n > 1\) and so

   \begin{equation*} \frac{\ln n}{n} > \frac{1}{n}\text{.} \end{equation*}

   Thus the series

   \begin{equation*} \sum_{n=0}^{\infty} \frac{\ln n}{n} \end{equation*}

   diverges by comparison to the harmonic series. To determine whether the original series converges conditionally or diverges, we apply the Alternating Series Test: Let

   \begin{equation*} a_n = \frac{\ln n}{n}, \ \ \text{ and } \ \ f(x) = \frac{\ln x}{x}\text{.} \end{equation*}

   Then

   \begin{equation*} f'(x) = \diff{}{x} \frac{\ln x}{x} = \frac{1-\ln x}{x^2}\text{,} \end{equation*}

   and so \(f'(x) \lt 0\) for \(x > e\text{.}\) Thus the sequence \(\{a_n\}_{n=1}^{\infty}\) is decreasing for \(n \geq 4\text{.}\) Additionally, we see that

   \begin{equation*} \lim_{n\to\infty} a_n = \lim_{n\to\infty} \frac{\ln n}{n} = \lim_{n\to\infty} \frac{1/n}{1} = 0\text{.} \end{equation*}

   Therefore,

   \begin{equation*} \lim_{n\to\infty} \frac{\ln n}{n} \end{equation*}

   is conditionally convergent.

4. \(\ds\sum_{n=1}^\infty (-1)^{n-1} {\ln n\over n^3}\)

   Answer
   converges absolutely
   Solution
   We first notice that

   \begin{equation*} a_n = \frac{\ln(n)}{n^3} \geq 0, \end{equation*}

   and

   \begin{equation*} \lim_{n\to\infty} a_n \Heq \lim_{n\to\infty} \frac{1}{3n^3} = 0, \end{equation*}

   and

   \begin{equation*} \diff{}{x} \frac{\ln(x)}{x} = \frac{1-3\ln(x)}{x^4} \lt 0 \quad \text{for } x\in (e^{1/3}, \infty). \end{equation*}

   Hence, by the Alternating Series Test, the series converges. Furthermore, the corresponding series

   \begin{equation*} \sum_{n=1}^{\infty} \frac{\ln (n)}{n^3} \end{equation*}

   converges by comparison with the convergent \(p\)-series \(\sum \frac{1}{n^2}\text{,}\) since \(\ln (n) \lt n \text{.}\) Therefore, the alternating series converges absolutely.

5. \(\ds\sum_{n=2}^\infty (-1)^n{1\over \ln n}\)

   Answer
   converges conditionally
   Solution

   We first analyze the corresponding series of absolute values,

   \begin{equation*} \sum_{n=2}^{\infty} \frac{1}{\ln n}\text{.} \end{equation*}

   For \(n \geq 2\text{,}\) we have that

   \begin{equation*} \frac{1}{\ln n} > \frac{1}{n}\text{,} \end{equation*}

   and so the series diverges (since the harmonic series is divergent). Therefore, the original series either diverges or is conditionally convergent. We apply the Alternating Series Test:

   \begin{equation*} \lim_{n\to\infty} \frac{1}{\ln n} = 0\text{,} \end{equation*}

   and for \(n \geq 2\text{,}\)

   \begin{equation*} \ln (n+1) > \ln n \implies \frac{1}{\ln (n+1)} \lt \frac{1}{\ln n} \end{equation*}

   and also \(\ln n \gt 0\text{.}\) Thus, the series

   \begin{equation*} \sum_{n=2}^{\infty} \frac{(-1)^n}{\ln n} \end{equation*}

   is conditionally convergent.

6. \(\ds\sum_{n=0}^\infty (-1)^{n} {3^n\over 2^n+5^n}\)

   Answer
   converges absolutely
   Solution
   We first notice that the corresponding series

   \begin{equation*} \sum_{n=1}^{\infty} \frac{3^n}{2^n+5^n} \end{equation*}

   converges by comparison with the convergent geometric series \(\sum \left(\frac{3}{5}\right)^n\text{.}\) We now check the conditions for the Alternating Series Test:

   \begin{equation*} a_n = \frac{3^n}{2^n+5^n} \geq 0, \end{equation*}

   and

   \begin{equation*} \frac{3^{n+1}}{2^n+5^n} \leq \frac{2^{n+1} + 5^{n+1}}{3^n} \implies a_{n+1} \leq a_n, \end{equation*}

   and

   \begin{equation*} \lim_{n\to\infty} a_n = \lim_{n\to\infty} \frac{3^n}{2^n+5^n} = 0. \end{equation*}

   Hence, the series converges absolutely.

7. \(\ds\sum_{n=0}^\infty (-1)^{n} {3^n\over 2^n+3^n}\)

   Answer
   diverges
   Solution
   We have that

   \begin{equation*} \lim_{n\to\infty} \frac{3^n}{2^n+3^n} = \lim_{n\to\infty} \frac{1}{\left(\frac{2}{3}\right)^n+1} = 1, \end{equation*}

   and so by the \(n\)-th Term Test, the series diverges.

8. \(\ds\sum_{n=1}^\infty (-1)^{n-1} {\arctan n\over n}\)

   Answer
   converges conditionally
   Solution
   We first notice that the corresponding series

   \begin{equation*} \sum_{n=1}^{\infty} \frac{\arctan(n)}{n} \end{equation*}

   diverges since

   \begin{equation*} \frac{\arctan(n)}{n} \geq \frac{1}{n} \end{equation*}

   for all \(n \gt \tan(1).\) Therefore, if the given series converges, it does so conditionally. We now use the Alternating Series Test:

   \begin{equation*} a_n = \frac{\arctan(n)}{n} \geq 0, \end{equation*}

   and

   \begin{equation*} \lim_{n\to\infty} \frac{\arctan(n)}{n} \Heq \lim_{n\to\infty} \frac{1/(n^2+1)}{1}= 0. \end{equation*}

   Now let \(f(x) = \frac{\arctan(x)}{x}\text{.}\) Since

   \begin{equation*} f'(x) = \frac{x-\arctan(x) (x^2+1)}{x^2(x^2+1)} \lt 0 \end{equation*}

   for all \(x \gt 0\text{,}\) the corresponding sequence \(\{a_n\}\) is decreasing. Therefore, the series converges conditionally.