For the convergence tests of series, the $n$-th term test should be the first. If the limit as $n$ approaches infinity is not zero, the series diverges. Otherwise, other convergence tests should be applied after that. For simple tests, the following techniques may be useful even without proof.

A. Limit Comparison

Assume that $\sum b_n$ is already known about if it diverges or converges. To test $\sum a_n$, let $\displaystyle \lim_{n \to \infty} a_n / b_n = k$.

• If $k > 0$, both $\sum a_n$ and $\sum b_n$ converge or diverge together.
• If $k = 0$, $\sum a_n$ converges if $\sum b_n$ converge.
• If $k = \infty$, $\sum a_n$ diverges if $\sum b_n$ diverges.

B. Ratio — Useful When the Series includes $2^n$, $n!$, or $n^n$ Term

When $R = \displaystyle \lim_{n \to \infty} a_{n+1} / a_n$,

• If $0 < R < 1$, $\sum a_n$ converges.
• If $R > 1$, $\sum a_n$ diverges.
• If $R = 1$, this test is inconclusive.

C. Cauchy — Useful With a Power Series

When $r = \displaystyle \lim_{n \to \infty} \sqrt{n}{a_n}$,

• If $r < 1$, $\sum a_n$ converges.
• If $r > 1$, $\sum a_n$ diverges.
• If $r = 1$, this test is inconclusive.

D. Intuition

• $\displaystyle \sum_{n=1}^{\infty} \cfrac{a_n \pm c_n}{b_n} \approx \displaystyle \sum_{n=1}^{\infty} \cfrac{a_n}{b_n} \;$ if $a_n \gg c_n$.
• $\displaystyle \sum_{n=1}^{\infty} \cfrac{a_n}{b_n} \approx \displaystyle \sum_{n=1}^{\infty} \cfrac{1}{b_n} \;$ if $a_n \ll b_n$.
• $\displaystyle \sum_{n=1}^{\infty} \ln (1 + a_n) \approx \displaystyle \sum_{n=1}^{\infty} a_n$.
• $\sin f(n)$, $\tan f(n)$, $\sin^{-1} f(n)$, or $\tan^{-1} f(n)$ can be replaced by $f(n)$.
• $\displaystyle \sum_{n=1}^{\infty} \cfrac{1}{n} \sin \cfrac{1}{n} \approx \displaystyle \sum_{n=1}^{\infty} \cfrac{1}{n} \cdot \cfrac{1}{n} \implies$ converge
• $\displaystyle \sum_{n=1}^{\infty} \cfrac{1}{n} \sin^{-1} \cfrac{1}{n} \approx \displaystyle \sum_{n=1}^{\infty} \cfrac{1}{n} \cdot \cfrac{1}{n} \implies$ converge
• $\displaystyle \sum_{n=1}^{\infty} \tan \cfrac{1}{n^2} \approx \displaystyle \sum_{n=1}^{\infty} \cfrac{1}{n^2} \implies$ converge
• $\displaystyle \sum_{n=1}^{\infty} \tan^{-1} \cfrac{1}{n^2} \approx \displaystyle \sum_{n=1}^{\infty} \cfrac{1}{n^2} \implies$ converge
• $\displaystyle \sum_{n=2}^{\infty} \cfrac{1}{(\ln n)^n} \implies$ converge
• $\displaystyle \sum_{n=2}^{\infty} \cfrac{1}{(\ln n)^k} \quad (k \text{ is positive}) \implies$ diverge
• $\displaystyle \sum \cfrac{(\ln f(n))'}{(\ln f(n))^p} \implies$ converge if $p > 1$, and diverge if $0 < p \leq 1$
• Comparison of series
• $\cfrac{1}{n^n} < \cfrac{1}{n!} < \cfrac{1}{e^n}, \; \cfrac{1}{2^n} < \cfrac{1}{n^2} \implies$ converge
• $\cfrac{1}{n} < \cfrac{1}{\ln n} < \vert \sin n \vert, \; \vert \sin^{-1} n \vert, \; \text{constant} < \ln n < n < e^n, \; 2^n < n! < n^n \implies$ diverge

The radius of Convergence is half of the interval of convergence for a power series. For a power series $\displaystyle \sum_{n=0}^{\infty} a_n f(x)^n$, the radius of Convergence $r$ is \begin{aligned} r = \lim_{n \to \infty} \left \vert \cfrac{a_n}{a_{n+1}} \right \vert \end{aligned}

In other words, the interval of convergence can be set $\vert f(x) \vert < r$. The test about boundaries of this interval should be determined after putting $r$ into the power series. There are techniques to find $r$ easily depending on the form of $a_n$ as follows.

$a_n = n$-th degree polynomial, or $(-1)^n$$r = 1$
$a_n = \ln n, \sqrt{n}, \sin n$$r = 1$
$a_n = b^n$$r = \cfrac{1}{b}$
$a_n = n!$$r = 0$
$a_n = \cfrac{(n!)^b}{(bn)!}$$r = b^b$
$a_n = \cfrac{n!}{n^n}$$r = e$
$a_n = \left( 1 + \cfrac{1}{n} \right)$$r = 1$

1. Integral

Given a monotone decreasing sequence $\langle a_n \rangle$, $\sum a_n$ converges if and only if $\displaystyle \int_m^{\infty} a_n dn$ is finite for $m \geq n$. If the integral diverges, then the series diverges as well.

[Proof] Consider $a_n$ as a curve. Then, the total area of rectangles below this curve is $a_2 + a_3 + \cdots$. Also, the total area of rectangles above this curve is $a_1 + a_2 + \cdots$. Since the difference between the two areas is $a_1$, this does not affect the convergence test. As such, the following inequalities hold. \begin{aligned} a_2 + a_3 + \cdots < \int_m^{\infty} a_n dn < a_1 + a_2 + \cdots \end{aligned}

Since the integral is between the two areas, the convergence test depends on whether the integral is finite.

2. Partial Sum

For two sequences $\langle a_n \rangle$, $\langle b_n \rangle$, and the partial sum $s_n = \sum_{k = 1}^{n} a_k$, the following equation holds. \begin{aligned} \sum_{k = 1}^{n} a_k b_k = \sum_{k = 1}^{n} s_k (b_k - b_{k + 1}) + s_n b_{n + 1} \end{aligned}

[Proof] For convenience, let $s_0 = 0$. \begin{aligned} \sum_{k = 1}^{n} a_k b_k &= \sum_{k = 1}^{n} (s_k - s_{k - 1}) b_k = \sum_{k = 1}^{n} s_k b_k - \sum_{k = 1}^{n} s_{k - 1} b_k \\ &= \sum_{k = 1}^{n} s_k b_k - \sum_{k = 1}^{n} s_k b_{k + 1} + s_n b_{n + 1} \\ &= \sum_{k = 1}^{n} s_k (b_k - b_{k + 1}) + s_n b_{n + 1} \end{aligned}

Therefore, $\sum a_k b_k$ converges if and only if $\sum s_k (b_k - b_{k + 1})$ and the sequence $\langle s_n b_{n + 1} \rangle$ converge.

3. Dirichlet

If the partial sum $s_n = \sum_{k = 1}^{n} a_k$ of a sequence $\langle a_n \rangle$ is bounded, and $\lim_{n \to \infty} b_n = 0$ and $b_n \geq b_{n + 1}$ for another sequence $\langle b_n \rangle$, then $\sum_{n = 1}^{\infty} a_n b_n$ converges.

[Proof] Since $b_n \geq b_{n + 1}$, \begin{aligned} \sum_{k = 1}^{n} \left| b_k - b_{k + 1} \right| &= \left| b_1 - b_2 \right| + \left| b_2 - b_3 \right| + \cdots + \left| b_n - b_{n + 1} \right| \\ &= (b_1 - b_2) + (b_2 - b_3) + \cdots + (b_n - b_{n + 1}) = b_1 - b_{n + 1} \\\\ \Longrightarrow \sum_{n = 1}^{\infty} \left| b_n - b_{n + 1} \right| &= b_1 \end{aligned}

Therefore, $\sum_{n = 1}^{\infty} a_n b_n$ converges as well by the partial sum theorem.

4. Abel

If $\sum_{n = 1}^{\infty} a_n$ converges and a monotonic sequence $\langle b_n \rangle$ is bounded, then $\sum_{n = 1}^{\infty} a_n b_n$ converges.

[Proof] If $\langle b_n \rangle$ is increasing, \begin{aligned} \sum_{k = 1}^{n} \left| b_k - b_{k + 1} \right| &= \left| b_1 - b_2 \right| + \left| b_2 - b_3 \right| + \cdots + \left| b_n - b_{n + 1} \right| \\ &= (b_2 - b_1) + (b_3 - b_2) + \cdots + (b_{n + 1} - b_n) = b_{n + 1} - b_1 \end{aligned}

If $\langle b_n \rangle$ is decreasing, \begin{aligned} \sum_{k = 1}^{n} \left| b_k - b_{k + 1} \right| &= \left| b_1 - b_2 \right| + \left| b_2 - b_3 \right| + \cdots + \left| b_n - b_{n + 1} \right| \\ &= (b_1 - b_2) + (b_2 - b_3) + \cdots + (b_n - b_{n + 1}) = b_1 - b_{n + 1} \end{aligned}

Therefore, $\sum_{n = 1}^{\infty} \vert b_n - b_{n + 1} \vert$ converges. Let $s_n = \sum_{k = 1}^{n} a_k$, then $s_n$ is bounded since $\sum_{n = 1}^{\infty} a_n$ converges, which means that $\sum_{n = 1}^{\infty} s_n (b_n - b_{n + 1})$ converges absolutely. In addition, $\langle b_n \rangle$ also converges by the monotone convergence theorem, which implies that $\langle s_n b_{n + 1} \rangle$ converges. Therefore, $\sum_{n = 1}^{\infty} a_n b_n$ converges as well by the partial sum theorem.

5. Kummer

Let two sequences of positive constants be $\langle a_n \rangle$ and $\langle b_n \rangle$, and $c_n = \cfrac{a_n b_n}{a_{n + 1}} - b_{n + 1}$.

• (i) If $\liminf\limits_{n \to \infty} c_n > 0$, then $\sum a_n$ converges.
• (ii) If $\limsup\limits_{n \to \infty} c_n < 0$ and $\sum \cfrac{1}{b_n}$ diverges, then $\sum a_n$ diverges.

[Proof] (i) Let $\rho = \liminf\limits_{n \to \infty} c_n$. Since $\rho/2 > 0$, for some natrual number $N$ such that $n > N$, \begin{aligned} c_n \geq \rho - \frac{\rho}{2} = \frac{\rho}{2} \end{aligned}

Since $a_{n + 1} > 0$ for $n > N$, then \begin{aligned} b_n a_n - b_{n + 1} a_{n + 1} \geq \frac{\rho}{2} a_{n + 1} \end{aligned}

By mathematical induction, for a natrual number $p$, the following inequalities hold. \begin{aligned} \frac{\rho}{2} a_{N + 1} &\leq b_N a_N - b_{N + 1} a_{N + 1} \\ \frac{\rho}{2} a_{N + 2} &\leq b_{N + 1} a_{N + 1} - b_{N + 2} a_{N + 2} \\ &\vdots \\ \frac{\rho}{2} a_{N + p} &\leq b_{N + p - 1} a_{N + p - 1} - b_{N + p} a_{N + p} \\\\ \Longrightarrow \frac{\rho}{2} (a_{N + 1} + &a_{N + 2} + \cdots + a_{N + p}) \leq b_N a_N - b_{N + p} a_{N + p} \end{aligned}

For $s_n = \sum_{k = 1}^{n} a_k$, \begin{aligned} s_{N + p} - s_N \leq \frac{2}{\rho} (b_N a_N - b_{N + p} a_{N + p}) \leq \frac{2}{\rho} b_N a_N \end{aligned}

This result implies that the sequence $\langle s_n \rangle$ is bounded. Therefore, $\sum a_n$ converges.

(ii) For some natrual number $N$ such that $n > N$, $c_n < 0$. That is, the sequence $\langle a_n b_n \rangle$ is increasing since $b_n a_n - b_{n + 1} a_{n + 1} < 0$ for $n > N$. Therefore, \begin{aligned} \sum_{k = 1}^{n} a_k \geq \sum_{k = 1}^{N} a_k + \sum_{k = N + 1}^{n} \frac{b_N a_N}{b_k} = \sum_{k = 1}^{N} a_k + b_N a_N \sum_{k = N + 1}^{n} \frac{1}{b_k} \end{aligned}

Accordingly, $\sum a_n$ diverges since $\sum \cfrac{1}{b_n}$ diverges.

6. Raabe

For a sequence of positive constants $\langle a_n \rangle$, $\sum a_n$ converges if $\rho > 1$ and $\sum a_n$ diverges if $\rho < 1$ given \begin{aligned} \rho = \lim\limits_{n \to \infty} n \left( \cfrac{a_n}{a_{n + 1}} - 1\right) \end{aligned}

[Proof] Let $b_n = n$. Applying Kummer’s test to two sequences $\langle a_n \rangle$ and $\langle b_n \rangle$, \begin{aligned} \lim\limits_{n \to \infty} \left( \cfrac{a_n b_n}{a_{n + 1}} - b_{n + 1}\right) = \rho - 1 \end{aligned}

Therefore, $\sum a_n$ converges if $\rho - 1 > 0$ and $\sum a_n$ diverges if $\rho - 1 < 0$

7. Bertrand

Let $\langle a_n \rangle$ be a sequence of positive constants and $N$ be a natural number such that $N \geq 2$.

• (i) $\sum a_n$ converges if the following inequality holds for a real number $A > 1$ and $n \geq N$.
\begin{aligned} \cfrac{a_n}{a_{n + 1}} \geq 1 + \cfrac{1}{n} + \cfrac{A}{n \ln n} \end{aligned}
• (ii) $\sum a_n$ diverges if the following inequality holds for $n \geq N$.
\begin{aligned} \cfrac{a_n}{a_{n + 1}} \leq 1 + \cfrac{1}{n} + \cfrac{1}{n \ln n} \end{aligned}

[Proof] (i) From the definition of Euler’s number, \begin{aligned} \left( 1 + \frac{1}{n} \right)^n < e \iff n \ln \left( 1 + \frac{1}{n} \right) < 1 \iff \ln \left( \frac{n + 1}{n} \right) < \frac{1}{n} \iff 0 < \ln \left( \frac{n}{n + 1} \right) + \frac{1}{n} \end{aligned}

Now, reforming the original inequality, \begin{aligned} \cfrac{a_n}{a_{n + 1}} n \ln n &\geq (n + 1)\ln n + A \\ \Longrightarrow \cfrac{a_n}{a_{n + 1}} n \ln n - (n + 1) \ln (n + 1) &\geq (n + 1)\ln \left( \frac{n}{n + 1} \right) + A > -\frac{n + 1}{n} + A \end{aligned}

Since $A - 1 > 0$, for some natrual number $N_1$ such that $n \geq N_1$, \begin{aligned} \frac{1}{n} < \frac{1}{2} (A - 1) \end{aligned}

Thus, when $n \geq \max(N, N_1)$, \begin{aligned} \cfrac{a_n}{a_{n + 1}} n \ln n - (n + 1) \ln (n + 1) > -\frac{n + 1}{n} + A = -\frac{1}{n} + A - 1 > \frac{1}{2} (A - 1) > 0 \end{aligned}

Therefore, by Kummer’s test, $\sum a_n$ converges.

(ii) From the definition of Euler’s number, \begin{aligned} \left( 1 - \frac{1}{n} \right)^n < \frac{1}{e} \iff \left( 1 - \frac{1}{n} \right)^n < \left( 1 - \frac{1}{n + 1} \right)^{n + 1} < \frac{1}{e} \iff (n + 1) \ln \left( \frac{n}{n + 1} \right) < -1 \end{aligned}

Now, reforming the original inequality, \begin{aligned} \cfrac{a_n}{a_{n + 1}} n \ln n &\leq (n + 1)\ln n + 1 \\ \Longrightarrow \cfrac{a_n}{a_{n + 1}} n \ln n - (n + 1) \ln (n + 1) &\leq (n + 1)\ln \left( \frac{n}{n + 1} \right) + 1 < -1 + 1 = 0 \end{aligned}

Since $\sum_{n = 2}^{\infty} 1/(n \ln n)$ diverges, $\sum a_n$ diverges by Kummer’s test.

8. Gauss

Let $\langle a_n \rangle$ be a sequence of positive constants. $\sum a_n$ converges if and only if $A > 1$ if the following conditions hold.

• For a sequence $\langle v_n \rangle$, $\vert v_n \vert \leq C / n^{1 + k}$ where $k$ and $C$ are positive real numbers and $M$ is a natural number such that $n \geq M$.
• For some natrual number $N$ such that $n \geq N$, $\cfrac{a_n}{a_{n + 1}} = 1 + \cfrac{A}{n} + v_n$.

[Proof] By the assumption, for $n \geq N$, \begin{aligned} n \left( \cfrac{a_n}{a_{n + 1}} - 1 \right) = A + n v_n \end{aligned}

Also, $\vert n v_n \vert \leq C / n^{k}$ since $\vert v_n \vert \leq C / n^{1 + k}$ for $n \geq M$. When $k > 0$, $\lim\limits_{n \to \infty} 1 / n^k = 0$, then $\lim\limits_{n \to \infty} n v_n = 0$. Therefore, \begin{aligned} \lim_{n \to \infty} n \left( \cfrac{a_n}{a_{n + 1}} - 1 \right) = A + \lim_{n \to \infty} n v_n = A \end{aligned}

By Raabe’s test, $\sum a_n$ converges if $A > 1$ and $\sum a_n$ diverges if $A < 1$. Now, consider the case of $A = 1$. Let $N_1 = \max(N, M)$. Reforming the original equation for $n \geq N_1$, \begin{aligned} \cfrac{a_n}{a_{n + 1}} = 1 + \frac{1}{n} + \frac{v_n n \ln n}{n \ln n} \end{aligned}

For $n \geq N_1$, $\vert v_n n \ln n \vert \leq C \ln n / n^k$. Also, since $\lim\limits_{n \to \infty} \ln n / n^k = 0$ for $k > 0$, $\lim\limits_{n \to \infty} v_n n \ln n = 0$. Therefore, for some natrual number $N_2$ such that $n \geq N_2$, $\vert v_n n \ln n \vert \leq 1$. Thus, when $n = \max(N_1, N_2)$, \begin{aligned} \cfrac{a_n}{a_{n + 1}} \leq 1 + \frac{1}{n} + \frac{1}{n \ln n} \end{aligned}

By Bertrand’s test, $\sum a_n$ diverges.

9. Simplified Gauss

From Gauss’s test, consider that $v_n = r_n / n^2$ and $k = 1$. Let $\langle a_n \rangle$ be a sequence of positive constants, and $\langle r_n \rangle$ be a bounded sequence. $\sum a_n$ converges if and only if $A > 1$ if the following condition holds. \begin{aligned} \cfrac{a_n}{a_{n + 1}} = 1 + \frac{A}{n} + \frac{r_n}{n^2} \end{aligned}

[Proof] It is trivial by Gauss’s test.

10. Binomial series

For a real number $\beta$ and a nonnegative integer $n$, let $a_n = \binom{\beta}{n}$.

• (i) $\sum\limits_{n = 0}^{\infty} a_n x^n$ converges if $\vert x \vert < 1$.
• (ii) $\sum\limits_{n = 0}^{\infty} a_n x^n$ converges absolutely if $\beta > 0$ and $x = \pm 1$.
• (iii) $\sum\limits_{n = 0}^{\infty} a_n x^n$ diverges if $\beta < 0$ and $x = -1$.
• (iv) $\sum\limits_{n = 0}^{\infty} a_n x^n$ diverges if $\beta \leq -1$ and $x = 1$.
• (v) $\sum\limits_{n = 0}^{\infty} a_n x^n$ converges conditionally if $-1 < \beta < 0$ and $x = 1$.

[Proof] (i) By the ratio test, $\sum_{n = 0}^{\infty} a_n x^n$ converges considering the following, \begin{aligned} \cfrac{a_n}{a_{n + 1}} &= \cfrac{\beta !}{(\beta - n)! n!} \cfrac{(\beta - n - 1)!(n + 1)!}{\beta !} = \cfrac{n + 1}{\beta - n} \\\\ \Longrightarrow &\lim_{n \to \infty} \left \vert \cfrac{a_{n + 1} x^{n + 1}}{a_n x^n} \right \vert = \lim_{n \to \infty} \left \vert \cfrac{(\beta - n)x}{n + 1} \right \vert = \vert x \vert \end{aligned}

(ii) Let $n > \beta$, then \begin{aligned} \left \vert \cfrac{a_n}{a_{n + 1}} \right \vert &= \cfrac{n + 1}{n - \beta} = 1 + \cfrac{1 + \beta}{n - \beta} = 1 + \cfrac{1 + \beta}{n} \left( 1 + \cfrac{\beta}{n - \beta} \right) = 1 + \cfrac{1 + \beta}{n} + \cfrac{\beta(1 + \beta)}{n(n - \beta)} \\\\ &= 1 + \cfrac{1 + \beta}{n} + \cfrac{1}{n^2} \left( \cfrac{n \beta (1 + \beta)}{n - \beta} \right) \end{aligned}

Let $A = 1 + \beta$ and $A_n = n \beta (1 + \beta) / (n - \beta)$. For $n > \beta$, \begin{aligned} \vert A_n \vert = \left \vert \cfrac{\beta (1 + \beta)}{1 - \cfrac{\beta}{n}} \right \vert \leq \begin{cases} \cfrac{\vert \beta (1 + \beta) \vert}{1 - \cfrac{\beta}{\lfloor \beta \rfloor + 1}} \qquad (\beta > 0) \\\\ \vert \beta (1 + \beta) \vert \qquad (\beta < 0) \end{cases} \end{aligned}

So, a sequence $\langle A_n \rangle$ is bounded. By Gauss’s test, $\sum\limits_{n = 0}^{\infty} a_n x^n$ converges absolutely if $x = \pm 1$ and $A = 1 + \beta > 1$, which is $\beta > 0$.

(iii) Since $a_n x^n = a_n(-1)^n = \vert a_n \vert$, $\sum\limits_{n = 0}^{\infty} a_n x^n$ diverges.

(iv) $\sum\limits_{n = 0}^{\infty} a_n x^n$ diverges since the sequence $\langle a_n \rangle$ does not converge to zero considering the following, \begin{aligned} \vert a_n \vert &= \left \vert \cfrac{\beta !}{(\beta - n)! n!} \right \vert = \left \vert \cfrac{\beta (\beta - 1) \cdots (\beta - (n - 1))}{1 \cdot 2 \cdots \cdot n} \right \vert \\\\ &= \cfrac{\vert\beta\vert (\vert\beta\vert + 1) \cdots (\vert\beta\vert + (n - 1))}{1 \cdot 2 \cdots \cdot n} \geq 1 \end{aligned}

(v) Since $0 < -\beta < 1$, $k < -\beta + k < k + 1$. Then, \begin{aligned} a_n = \cfrac{\beta (\beta - 1) \cdots (\beta - (n - 1))}{1 \cdot 2 \cdots \cdot n} = (-1)^n \cfrac{(-\beta) (-\beta + 1) \cdots (-\beta + (n - 1))}{1 \cdot 2 \cdots \cdot n} \end{aligned}

Let $a_n = (-1)^n b_n$ for a sequence $\langle b_n \rangle$ such that $b_0 = 1$. Then, $0 < b_n < 1$ since $k < -\beta + k < k + 1$. Also, $\langle b_n \rangle$ is the sequence of positive numbers strictly decreasing since \begin{aligned} \cfrac{b_{n + 1}}{b_n} = \cfrac{(-\beta) (-\beta + 1) \cdots (-\beta + n)}{(n + 1)!} \cfrac{n!}{(-\beta) (-\beta + 1) \cdots (-\beta + (n - 1))} = \cfrac{-\beta + n}{n + 1} < 1 \end{aligned}

Considering $\ln t < t - 1$ for $0 < t < 1$, \begin{aligned} b_n &= \cfrac{b_n}{b_{n - 1}}\cfrac{b_{n - 1}}{b_{n - 2}} \cdots \cfrac{b_2}{b_1}b_1 \\\\ \implies \ln b_n &= \sum_{k = 1}^n \ln \cfrac{b_k}{b_{k - 1}} < \sum_{k = 1}^n \left( \cfrac{b_k}{b_{k - 1}} - 1 \right) = \sum_{k = 1}^n \left( \cfrac{-\beta + k - 1}{k} - 1 \right) = (-\beta - 1) \sum_{k = 1}^n \cfrac{1}{k} \end{aligned}

Since $-\beta - 1 < 0$ and $\sum\limits_{k = 1}^{\infty} 1/k = \infty$, $\lim\limits_{n \to \infty} \ln b_n = -\infty$ and $\lim\limits_{n \to \infty} b_n = 0$. By alternating series test, $\sum a_n x^n = \sum a_n = \sum (-1)^n b_n$ converges. Meanwhile, $\sum \vert a_n \vert$ diverges if $\beta < 0$. Therefore, $\sum\limits_{n = 0}^{\infty} a_n x^n$ converges conditionally.