Convergence and Divergence Theorems for Series

We will now look at some other very important convergence and divergence theorems apart from the The Divergence Theorem for Series.

Theorem 1: The series $\sum_{n=1}^{\infty} a_n$ is convergent if and only if for any natural number $N \in \mathbb{N}$ the series $\sum_{n=N}^{\infty} a_n$ is also convergent.

• Proof of Theorem 1: $\Rightarrow$ Suppose that $\sum_{n=1}^{\infty} a_n$ is convergent, that is $\lim_{n \to \infty} s_n = L$ for some $L \in \mathbb{R}$. We will proceed to prove this with mathematical induction. Let $P(N)$ be the statement that $\sum_{n=N}^{\infty} a_n$ is convergent.

• For our base step, if $N = 1$, then we already know that $\sum_{n=1}^{\infty} a_n$ is convergent.

• Now suppose that for some $k \in \mathbb{N}$, the statement $P(k) : \sum_{n=k}^{\infty} a_n$ is convergent to $S$ . We want to show that the statement $P(k+1) : \sum_{n=k+1}^{\infty} a_n$ is also convergent. Now notice that:

(1)

\begin{align} \sum_{n=k+1}^{\infty} a_n = \sum_{n=k}^{\infty} a_n - a_k \end{align}

• Since $\sum_{n=k}^{\infty} a_n$ is convergent to some sum $S$ by the induction hypothesis, then $\sum_{n=k+1}^{\infty} a_n$ is convergent to $S - a_k$. Therefore $P(k+1)$ is true.

• Therefore for any $n \in \mathbb{N}$, if $\sum_{n=1}^{\infty} a_n$ is convergent then $\sum_{n=N}^{\infty} a_n$ is convergent.

• $\Leftarrow$ Suppose that the series $\sum_{n=N}^{\infty} a_n$ is convergent for any $n \in \mathbb{N}$. Then if $N = 1$, $\sum_{n=1}^{\infty} a_n$ is convergent, and we are done the proof $\blacksquare$

Theorem 2: If the sequence $\{ a_n \}$ is ultimately positive and the sequence of partial sums $\{ s_n \}$ is bounded above then the series $\sum_{n=1}^{\infty} a_n$ converges. If the sequence of partial sums is not bounded above then $\sum_{n=1}^{\infty} a_n$ diverges to $\infty$.

• Proof of Theorem 2: Suppose that $\{ a_n \}$ is ultimately positive, that is for some $N \in \mathbb{N}$, if $n ≥ N$ then $a_n > 0$. Let $\{ s_n \}$ be the sequence of partial sums that is bounded above by $M \in \mathbb{R}$ such that $a_1 + a_2 + ... + a_n = s_n ≤ M$ for all $n \in \mathbb{N}$.

• We note that $\{ s_n \}$ must then be ultimately increasing. Consider $s_N = a_1 + a_2 + ... + a_N$ and $s_{N+1} = a_1 + a_2 + ... + a_N + a_{N+1}$. Now we know that for some $N \in \mathbb{N}$ if $n ≥ N$ then $a_n > 0$. So therefore $a_{N+1} > 0$ and thus $s_N < s_{N+1}$.

• Since $\{ s_n \}$ is ultimately increasing and bounded above, the sequence $\{ s_n \}$ must be convergent by the The Monotonic Sequence Theorem for Convergence and therefore $\sum_{n=1}^{\infty} a_n$ is convergent. $\blacksquare$

Example 1

Show that the series $\sum_{n=1}^{\infty} \frac{1 - n}{2 + n}$ is divergent.

We note that $\lim_{n \to \infty} \frac{1 - n}{2 + n} = \lim_{n \to \infty} \frac{\frac{1}{n} - 1}{\frac{2}{n} + 1} = -1$. Since $L < 0$, we reason that the series $\sum_{n=1}^{\infty} \frac{1 - n}{2 + n}$ is divergent to negative infinity.

Example 2

Show that the series $\sum_{n=1}^{\infty} \frac{n^2 - 1}{n}$ is divergent.

We note that $\lim_{n \to \infty} \frac{n^2 - 1}{n} = \lim_{n \to \infty} n - \frac{1}{n} = \infty$. So the series $\sum_{n=1}^{\infty} \frac{n^2 - 1}{n}$ is divergent to infinity.