Applying Convergence and Divergence Tests for Series

Applying Convergence and Divergence Tests for Series

Thus far we have looked at various techniques to determine whether a series converges or diverges. In reality though, we aren't often told which technique to apply specifically and instead we will have to apply our intuition to determine the best technique to try. The following list summarizes all of the convergence and divergence theorems we have looked into thus far.

  • Geometric Series Test: $\sum_{n=1}^{\infty} ar^{n-1}$ converges (to $\frac{a}{1 - r}$) if $\mid r \mid < 1$ and diverges if $\mid r \mid ≥ 1$.
  • Divergence Test: If $\lim_{n \to \infty} a_n \neq 0$, then $\sum_{n=1}^{\infty} a_n$ diverges.
  • Integral Test: If $f(n) = a_n$ for some positive decreasing continuous function on the interval $[N, \infty)$ then if $\int_{N}^{\infty} f(x) \: dx$ converges then $\sum_{n=N}^{\infty} a_n$ converges. If $\int_{N}^{\infty} f(x) \: dx$ diverges then $\sum_{n=N}^{\infty} a_n$ diverges to infinity.
  • p-Series Test: $\sum_{n=1}^{\infty} \frac{1}{n^p}$ is convergent if $p > 1$ and divergent if $p ≤ 1$.
  • Comparison Test: For two ultimately positive sequences $\{ a_n \}, \{ b_n \}$ such that for some $N \in \mathbb{N}$, if $n ≥ N$ then $0 ≤ a_n ≤ b_n$, then if $\sum_{n=1}^{\infty} b_n$ is convergent then $\sum_{n=1}^{\infty} a_n$ is convergent. If $\sum_{n=1}^{\infty} a_n$ is divergent then $\sum_{n=1}^{\infty} b_n$ is divergent.
  • Limit Comparison Test: For two positive sequences $\{ a_n \}, \{ b_n \}$, if $\lim_{n \to \infty} \frac{a_n}{b_n} = L$, if $0 < L < \infty$ then either both series converge or both series diverge. If $L = 0$ this test fails.
  • Ratio Test: If $\{ a_n \}$ is ultimately positive and $\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \rho$, if $0 ≤ \rho < 1$ then $\sum_{n=1}^{\infty} a_n$ is convergent. If $1 < \rho ≤ \infty$ then $\sum_{n=1}^{\infty} a_n$ is divergent. If $\rho = 1$ the test fails.
  • Root Test: If $\{ a_n \}$ is ultimately positive and $\lim_{n \to \infty} (a_n)^{1/n} = L$, if $0 ≤ L < 1$ then $\sum_{n=1}^{\infty} a_n$ is convergent. If $1 < L ≤ \infty$ then $\sum_{n=1}^{\infty} a_n$ is divergent. If $L = 1$ the test fails.
  • Alternating Series Test: If $a_na_{n+1} < 0$, $\mid a_{n+1} \mid ≤ \mid a_n \mid$, and $\lim_{n \to \infty} a_n = 0$, then $\sum_{n=1}^{\infty} a_n$ is convergent.

We will now look at some examples of applying these tests. More examples can be found on the following pages:

Example 1

Determine whether $\sum_{n=1}^{\infty} \frac{\sqrt{n}}{n^2 + n + 1}$ is convergent or divergent.

We note that as $n$ gets very large, $a_n$ behaves like $\frac{1}{n^{3/2}}$, and so we will apply the limit comparison test.

\begin{align} \lim_{n \to \infty} \frac{\frac{\sqrt{n}}{n^2 + n + 1}}{\frac{1}{n^{3/2}}} = \lim_{n \to \infty} \frac{n^2}{n^2 + n + 1} = 1 \end{align}

And therefore since $L \neq 0$, both series must either converge or diverge. We know by the p-Series test that $\sum_{n=1}^{\infty} \frac{1}{n^{3/2}}$ converges so $\sum_{n=1}^{\infty} \frac{\sqrt{n}}{n^2 + n + 1}$ must also converge.

Example 2

Determine whether $\sum_{n=1}^{\infty} \frac{1}{n \ln (n) \sqrt{\ln(\ln(n))}}$ is convergent or divergent.

For this example we should use the integral test since many of the other tests seem complicated. Let $f(x) = \frac{1}{x \ln (x) \sqrt{\ln(\ln(x))}}$. We note that $f$ is positive for $x ≥ 1$ and $f$ is decreasing and so let's evaluate the following integral

\begin{align} \quad \int_{1}^{\infty} \frac{1}{x \ln x \sqrt{\ln(\ln(x))}} \: dx = \lim_{b \to \infty} \int_{1}^{b} \frac{1}{x \ln x} \cdot \frac{1}{\sqrt{\ln(\ln x))}} \: dx \end{align}

Using substitution, let $u = \ln ( \ln x )$. Then $du = \frac{1}{x \ln x } \: dx$, and therefore

\begin{align} \quad \int \frac{1}{x \ln x} \cdot \frac{1}{\sqrt{\ln(\ln x))}} \: dx = \int \frac{1}{\sqrt{u}} \: du = \int u^{-1/2} \: du = 2\sqrt{u} = 2 \sqrt{\ln ( \ln x )} + C \end{align}

And so:

\begin{align} \quad \lim_{b \to \infty} \int_{1}^{b} \frac{1}{x \ln x} \cdot \frac{1}{\sqrt{\ln(\ln x))}} \: dx = \lim_{b \to \infty} 2 \sqrt{\ln ( \ln b )} - 2 \sqrt{\ln ( \ln 1 )} = \lim_{b \to \infty} 2 \sqrt{\ln ( \ln b )} = \infty \end{align}

Therefore the series $\sum_{n=1}^{\infty} \frac{1}{n \ln (n) \sqrt{\ln(\ln(n))}}$ is divergent.

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