Convergence Tests

Divergence test

$$\lim_{k \to \infty} u_k \neq 0 \implies \sum_{k=1}^n u_k\; \text{is diverging}$$

Direct Comparison Test

Let $0<u_k<v_k$. $$

$$\sum_{k=1}^\infty v_k\;\text{is converges} \implies \sum_{k=1}^\infty u_k\;\text{is converges}$$

Proof Hint

• Note that $\sum_{k=1}^n u_k$ and $\sum_{k=1}^n v_k$ are increasing
• Show that $\sum_{k=1}^\infty v_k$ converges to its supremum $v$ which is an upper bound of $\sum_{k=1}^n u_k$

Example

Proving the convergence of $\sum_{k=1}^\infty \frac{1}{k!}$, by using $k! \ge 2^{k-1}$ for all $k\ge 0$.

Limit Comparison Test

Let $0<u_k,v_k$ and $\lim_{n\to \infty}{\frac{u_n}{v_n}} =R$.

$$R \gt 0\implies \bigg(\sum_{n=1}^\infty u_n\;\text{is converging}\iff \sum_{n=1}^\infty v_n\;\text{is converging}\bigg)$$ $$R = 0\implies \bigg(\sum_{n=1}^\infty v_n\;\text{is converging} \implies \sum_{n=1}^\infty u_n\;\text{is converging}\bigg)$$ $$R = \infty\implies\bigg(\sum_{n=1}^\infty v_n\;\text{is diverging} \implies \sum_{n=1}^\infty u_n\;\text{is diverging}\bigg)$$

Proof Hint

Only possibilities are $R=0,R\gt 0, R=\infty$. $$

For $R \gt 0$: $$

• Consider limit definition with $\epsilon=\frac{L}{2}$
• Direct comparison test can be used for the 2 set of inequalities

For $R=0$: $$

• Consider limit definition with $\epsilon=1$
• Direct comparison test can be used now

For $R=\infty$: $$

• Consider limit definition with $M=1$
• Direct comparison test can be used now

Integral Test

Let $u(x) \gt 0$, decreasing and integrable on $[1,M]$ for all $M \gt 1$. Then:

$$\sum_{n=1}^\infty u_n\text{ is converging} \iff \int_1^\infty {u(x)\,\text{d}x}\text{ is converging}$$

Proof Hint

As $u(x)$ is decreasing, it is apparent that it is integrable. $$

Make use of this inequality:

$$s_n - u_1 \le \int_1^n {u(x)\,\text{d}x} \le s_n - u_n$$

For $\impliedby$: $$

• Note that $s_n$ is increasing
• Show that $s_n$ is bounded above by $\int_1^\infty {u(x)\,\text{d}x} + u_1$

For $\implies$: $$

• Define $F(n)=\int_1^n {u(x)\,\text{d}x}$
• Note that $F(n)$ is increasing
• Note that $\lim_{n\to\infty}{u_n} = 0$
• Show that $F(n)$ is bounded above by $\lim_{n\to\infty}{s_n}$

Note

$$\sum_{n=1}^\infty u_n\text{ is converging} \implies \lim_{k \to \infty} u_k = 0$$$$\int_1^\infty {u(x)\,\text{d}x}\text{ is converging} \implies \lim_{k \to \infty} u(k) = 0$$

Ratio Test

Let $u(x) \gt 0$ and $\lim_{n\to \infty}{\frac{u_{n+1}}{u_n}} =L$.

$$L \lt 1 \implies \sum_{n=1}^\infty u_n\;\text{is converging}$$ $$L \gt 1 \implies \sum_{n=1}^\infty u_n\;\text{is diverging}$$

Proof Hint

• Consider the limit definition with:
  • For the $L \lt 1$ case: $\epsilon=\frac{1}{2}(1-L)$
  • For the $L \gt 1$ case: $\epsilon=\frac{1}{2}(L-1)$
• Show that: $\frac{1}{2}(3L-1) \lt \frac{u_{k+1}}{u_k} \lt \frac{1}{2}(1+L)$
• Recursively simplify the inequality to reach $u_{N+1}$ which is a constant
• Use $\sum_{k=1}^\infty r^k$ is converging iff $r < 1$

Root Test

Let $u_n$ is a sequence and $\lim_{n\to \infty}{(\rvert u_n\rvert)^{1/n}} =L$.

$$L \lt 1 \implies \sum_{n=1}^\infty u_n\;\text{is absolutely converging}$$ $$(L \gt 1 \lor L=\infty) \implies \sum_{n=1}^\infty u_n\;\text{is diverging}$$

Proof Hint

Consider the limit definition with:

• For $L<1$: $\epsilon=\frac{1}{2}(1-L)$
• For $L > 1$: $\epsilon=\frac{1}{2}(L-1)$
• For $L=\infty$:$M>1$

Dirichlet’s test

Let:

• $b_n$ is a decreasing sequence, converging to $0$ and
• $a_n$ is a sequence and
• $A_n = \sum_{k=1}^n a_k$ is bounded

$$\implies \sum_{n=1}^\infty a_nb_n\; \text{is converging}$$