In this post we will continue to investigate more theorems of convergence tests.

Tests

D’Alembert Ratio Test

$\textbf{\small Theorem 1.17.:} $ Let $\sum_{n=1}^{\infty} a_n$ be positive series and let $l = \lim_{n\rightarrow \infty}\frac{a_{n+1}}{a_n}$. Then, series converges if $l < 1$, series diverges if $l > 1$, and gives no information if $l = 1$ or limit fails to exist.

$\textbf{\small Proof:} $ Let $l < 1$. We want to prove that $\sum_{n=1}^{\infty} a_n$ converges. Let us write what we know from assumption. Choose $\varepsilon > 0$ and $\forall n > N$ and $0 < (l + \varepsilon) < 1$:

$$ \begin{align} \left| \dfrac{a_{n+1}}{a_n} - l\right | &< \varepsilon \notag \\ l - \varepsilon < \dfrac{a_{n+1}}{a_n} &< l + \varepsilon \notag \\ a_n \cdot (l -\varepsilon) < a_{n+1} &< a_n \cdot (l + \varepsilon) \end{align} $$
$$ \begin{align} \left| \dfrac{a_{n+1}}{a_n} - l\right | &< \varepsilon \notag\\ l - \varepsilon < \dfrac{a_{n+1}}{a_n} &< l + \varepsilon \notag\\ a_n \cdot (l -\varepsilon) < a_{n+1} &< a_n \cdot (l + \varepsilon) \end{align} $$

Let us take the right-hand side, in other words $a_{n+1} < a_n \cdot (l + \varepsilon)$, $\forall n > N$ where $n = N+1, N+2, \dots$

So we can write,

$$ \begin{align} a_{N+2} < a_{N+1} \cdot (l+\varepsilon) \end{align} $$
$$ \begin{align} a_{N+2} < a_{N+1} \cdot (l+\varepsilon) \end{align} $$

using $(2)$ we can write $n = N+2$,

$$ \begin{aligned} a_{N+3} < a_{N+2} \cdot (l+\varepsilon) < a_{N+1} \cdot (l+\varepsilon)^2 \end{aligned} $$
$$ \begin{aligned} a_{N+3} < a_{N+2} \cdot (l+\varepsilon) < a_{N+1} \cdot (l+\varepsilon)^2 \end{aligned} $$

we apply this process up to $n = N+k$,

$$ \begin{aligned} a_{N+k} < a_{N+k-1} \cdot (l+\varepsilon) < a_{N+1} \cdot (l+\varepsilon)^{k-1} \end{aligned} $$
$$ \begin{aligned} a_{N+k} < a_{N+k-1} \cdot (l+\varepsilon) < a_{N+1} \cdot (l+\varepsilon)^{k-1} \end{aligned} $$

thus, we have

$$ \begin{align} 0 < a_{N+k} < a_{N+1} \cdot (l+\varepsilon)^{k-1} \end{align} $$
$$ \begin{align} 0 < a_{N+k} < a_{N+1} \cdot (l+\varepsilon)^{k-1} \end{align} $$

Now, consider geometric series $a_{N+1} \sum_{k=1}^{\infty} (l+\varepsilon)^{k-1}$. Observe that, it converges since $(l+\epsilon) < 1$ (see here).

Finally, using direct comparison test with inequality in $(3)$ and convergence of geometric series, we can say $\sum_{k=1}^{\infty} a_{N+k}$ converges.

Similarly, take the left-hand side of the $(1)$. Let us assume $l > 1$ so that $(l - \varepsilon) > 1$. We can write below inequality for all $n > N$ where $n = N+1, N+2, \dots$

$$ \begin{aligned} a_n \cdot (l - \varepsilon) < a_{n+1} \end{aligned} $$
$$ \begin{aligned} a_n \cdot (l - \varepsilon) < a_{n+1} \end{aligned} $$

Using the same procedure as above we obtain

$$ \begin{aligned} 0 < a_{N+1} \cdot (l - \varepsilon)^{k-1} < a_{N+k} \end{aligned} $$
$$ \begin{aligned} 0 < a_{N+1} \cdot (l - \varepsilon)^{k-1} < a_{N+k} \end{aligned} $$

Notice that, sum $a_{N+1}\sum_{k=1}^{\infty} (l-\varepsilon)^{k-1}$ diverges by geometric series test since $(l-\varepsilon) > 1$ and also $\sum_{k=1}^{\infty} a_{N+k}$ diverges by direct comparison test.

Case when $l = 1$ can be shown using examples. For instance, take harmonic series it is obvious that $\lim_{n \rightarrow \infty} \frac{a_{n+1}}{a_n} = 1$, and it diverges, however if you take $\sum_{n=1}^{\infty} \frac{1}{n^2}$ although it converges, the limit $\lim_{n\rightarrow \infty}\frac{a_{n+1}}{a_n}$ is equal to $1$. $\blacksquare$

Root Test

$\textbf{\small Theorem 1.18.:}$ Let $\sum_{n=1}^{\infty} a_n$ be positive series and let $l = \lim_{n\rightarrow \infty} \sqrt[n]{a_n}$. Then, series converges if $l < 1$, series diverges if $l > 1$, and gives no information if $l = 1$ or limit fails to exist.

$\textbf{\small Proof:}$ Let us write the limit more rigorously. $\forall \varepsilon > 0$, $\exists N = N(\varepsilon)$ such that $|\sqrt[n]{a_n} - l| < \varepsilon$, $\forall n > N$. Therefore,

$$ \begin{aligned} -\varepsilon < \sqrt[n]{a_n} - l &< \varepsilon\\ l - \varepsilon < \sqrt[n]{a_n} &< l + \varepsilon\\ (l - \varepsilon)^n < a_n &< (l + \varepsilon)^n\\ \end{aligned} $$
$$ \begin{aligned} -\varepsilon < \sqrt[n]{a_n} - l &< \varepsilon\\ l - \varepsilon < \sqrt[n]{a_n} &< l + \varepsilon\\ (l - \varepsilon)^n < a_n &< (l + \varepsilon)^n\\ \end{aligned} $$

using the same idea as in Theorem 1.17 we can obtain the proofs for cases $l > 1$ and $l < 1$. Let us examine the case $l=1$. Consider the sum $\sum_{n=1}^{\infty} \frac{1}{n}$ diverges, and $\lim_{n \rightarrow \infty} \frac{1}{\sqrt[n]{n}} = 1$. However, $\sum_{n=1}^{\infty} \frac{1}{n^2}$ converges, and $\lim_{n \rightarrow \infty} \frac{1}{\sqrt[n]{n^2}} = 1$. $\blacksquare$

Now we will present the proof of Raabe’s Test, which is used when the Root Test is inconclusive. In many analysis and calculus textbooks, the proof of this test is presented in a brief and somewhat opaque manner. In my view, however, mathematics is truly understandable only when every step is carefully explained and every idea is fully developed. Therefore, I will provide every detail and highlight every key trick involved in the proof, so that no questions remain by the end.

Raabe’s Test

$ \textbf{\small Theorem 1.19.:}$ Let $\sum_{n=1}^{\infty} a_n$ be positive series and let $l = \lim_{n\rightarrow \infty} n \left(\frac{a_n}{a_{n+1}} - 1\right)$. Then, series converges if $l > 1$, series diverges if $l < 1$, and gives no information if $l = 1$ or limit fails to exist.

$\textbf{\small Proof:}$ Let $l > 1$ and $(l - \varepsilon) > 1$. Let us investigate the limit:

$$ \begin{aligned} \left|n \cdot \left(\dfrac{a_n}{a_{n+1}} - 1\right) - l\right| &< \varepsilon\\ l - \varepsilon < n \cdot \left(\dfrac{a_n}{a_{n+1}} - 1\right) &< l+ \varepsilon\\ l - \varepsilon + n< n \cdot \dfrac{a_n}{a_{n+1}} &< l+ \varepsilon + n\\ \end{aligned} $$
$$ \begin{aligned} \left|n \cdot \left(\dfrac{a_n}{a_{n+1}} - 1\right) - l\right| &< \varepsilon\\ l - \varepsilon < n \cdot \left(\dfrac{a_n}{a_{n+1}} - 1\right) &< l+ \varepsilon\\ l - \varepsilon + n< n \cdot \dfrac{a_n}{a_{n+1}} &< l+ \varepsilon + n\\ \end{aligned} $$

Using the fact that $(l-\varepsilon)>1$ we get,

$$ \begin{align} \dfrac{n+1}{n} < \dfrac{l - \varepsilon + n}{n}< \dfrac{a_n}{a_{n+1}} \tag{1} \end{align} $$
$$ \begin{align} \dfrac{n+1}{n} < \dfrac{l - \varepsilon + n}{n}< \dfrac{a_n}{a_{n+1}} \tag{1} \end{align} $$

We know that for all $n\in \mathbb{N}$ the inequality $\frac{n+1}{n} > 1$ holds. Using this we see

$$ \begin{aligned} 1 < \dfrac{n+1}{n} &< \dfrac{a_n}{a_{n+1}} \\ 1 &< \dfrac{a_n}{a_{n+1}} \\ a_{n+1} &< a_n \\ \end{aligned} $$
$$ \begin{aligned} 1 < \dfrac{n+1}{n} &< \dfrac{a_n}{a_{n+1}} \\ 1 &< \dfrac{a_n}{a_{n+1}} \\ a_{n+1} &< a_n \\ \end{aligned} $$

This means that $a_n$ is a decreasing sequence. Now let us show $a_n$ has a lower bound. In order for us to show that we should use the inequality in $(1)$;

$$ \begin{aligned} a_2\left(\dfrac{2}{1}\right) < a_1 &\rightarrow a_2 < \dfrac{a_1}{2}\\ a_3\left(\dfrac{3}{2}\right) < a_2 &\rightarrow a_3 < \dfrac{a_1}{3}\\ a_4\left(\dfrac{4}{3}\right) < a_3 &\rightarrow a_4 < \dfrac{a_1}{4}\\ &\vdots\\ a_{n+1} \cdot n &< a_1\\ \end{aligned} $$
$$ \begin{aligned} a_2\left(\dfrac{2}{1}\right) < a_1 &\rightarrow a_2 < \dfrac{a_1}{2}\\ a_3\left(\dfrac{3}{2}\right) < a_2 &\rightarrow a_3 < \dfrac{a_1}{3}\\ a_4\left(\dfrac{4}{3}\right) < a_3 &\rightarrow a_4 < \dfrac{a_1}{4}\\ &\vdots\\ a_{n+1} \cdot n &< a_1\\ \end{aligned} $$

This shows us that $a_n\cdot n$ has a lower bound, namely $a_1$. Since we have shown $a_n$ is monotonic (descreasing) by monotone convergence theorem we conclude that $\lim_{n \rightarrow \infty} n \cdot a_n <\infty$.

Now, we will manipulate exact same inequality in order to get telescoping series

$$ \begin{aligned} &(l - \varepsilon + n)< n \cdot \dfrac{a_n}{a_{n+1}} < l+ \varepsilon + n\\ &(l - \varepsilon + n)a_{n+1} < n \cdot a_n \\ &(l - \varepsilon + n)a_{n+1} - (n+1)\cdot a_{n+1}< n \cdot a_n - (n+1)\cdot a_{n+1}\\ &(l - \varepsilon - 1)a_{n+1}< n \cdot a_n - (n+1)\cdot a_{n+1} \end{aligned} $$
$$ \begin{aligned} &(l - \varepsilon + n)< n \cdot \dfrac{a_n}{a_{n+1}} < l+ \varepsilon + n\\ &(l - \varepsilon + n)a_{n+1} < n \cdot a_n \\ &(l - \varepsilon + n)a_{n+1} - (n+1)\cdot a_{n+1}< n \cdot a_n - (n+1)\cdot a_{n+1} \end{aligned} $$

Consider partial sum $s_n = n \cdot a_n - (n+1) \cdot a_{n+1}$. Corresponding telescopic series $\sum_{n=1}^{\infty} n \cdot a_n - (n+1) \cdot a_{n+1}$ converges if and only if $\lim_{n\rightarrow \infty} (n+1) \cdot a_{n+1}$ converges.

We already know that $\lim_{n\rightarrow \infty} n \cdot a_{n}$ converges, therefore partial sum $s_n$ converges and all in all sum $\sum_{n=1}^{\infty} n \cdot a_n - (n+1) \cdot a_{n+1}$ converges.

Using direct comparison test with the inequality $(l - \varepsilon - 1)a_{n+1}< n \cdot a_n - (n+1)\cdot a_{n+1}$ we can say $(l-\varepsilon-1)\sum_{n=1}^{\infty}a_{n+1}$ converges.

Now, let us prove the divergence. Take $l < 1$ and choose $\varepsilon > 0$ so that $l - 1 + \varepsilon < 0$

$$ \begin{aligned} \left|n \cdot \left(\dfrac{a_n}{a_{n+1}} - 1\right) - l\right| &< \varepsilon\\ l - \varepsilon < n \cdot \left(\dfrac{a_n}{a_{n+1}} - 1\right) &< l+ \varepsilon\\ l - \varepsilon + n< n \cdot \dfrac{a_n}{a_{n+1}} &< l+ \varepsilon + n\\ n \cdot a_n &< (l + \varepsilon + n) \cdot a_{n+1} \\ n \cdot a_n - (n+1) \cdot a_{n+1} &< (l + \varepsilon + n) \cdot a_{n+1} - (n+1) \cdot a_{n+1} \\ n \cdot a_n - (n+1) \cdot a_{n+1} &< (l + \varepsilon -1) \cdot a_{n+1} < 0\\ n \cdot a_n - (n+1) \cdot a_{n+1} &< 0\\ a_n &< \dfrac{n+1}{n} a_{n+1} \end{aligned} $$
$$ \begin{aligned} \left|n \cdot \left(\dfrac{a_n}{a_{n+1}} - 1\right) - l\right| &< \varepsilon\\ l - \varepsilon < n \cdot \left(\dfrac{a_n}{a_{n+1}} - 1\right) &< l+ \varepsilon\\ l - \varepsilon + n< n \cdot \dfrac{a_n}{a_{n+1}} &< l+ \varepsilon + n\\ n \cdot a_n &< (l + \varepsilon + n) \cdot a_{n+1} \\ n \cdot a_n - (n+1) \cdot a_{n+1} &< (l + \varepsilon + n) \cdot a_{n+1} - (n+1) \cdot a_{n+1} \\ n \cdot a_n - (n+1) \cdot a_{n+1} &< (l + \varepsilon -1) \cdot a_{n+1} < 0\\ n \cdot a_n - (n+1) \cdot a_{n+1} &< 0\\ a_n &< \dfrac{n+1}{n} a_{n+1} \end{aligned} $$

It is obvious that $a_n$ is an increasing sequence. Take large $n > N$.

$$ \begin{aligned} 0 < (N)\cdot a_{N} < (n+1) \cdot a_{n+1} \end{aligned} $$
$$ \begin{aligned} 0 < \dfrac{N}{n+1}\cdot a_{N} < a_{n+1} \end{aligned} $$

Trivially, $a_N\cdot N\sum_{n=1}^{\infty}\frac{1}{n+1}$ diverges. By direct comparison test $\sum_{n=1}^{\infty} a_{n+1}$ diverges. $\blacksquare$

Cauchy Condensation Test

$ \textbf{\small Theorem 1.20.:}$ Let $\sum_{n=1}^{\infty} a_n$ be positive series and let $a_n$ be a decreasing sequence. Then, the series $\sum_{n=1}^{\infty} a_n$ and $\sum_{n=1}^{\infty} 2^na_{2^n}$ either both converge or diverge.

$\textbf{\small Proof:}$ Let $s_n = a_1+a_2+\dots + a_n$ be a partial sum. Let $t_n = 2a_1+4a_2+\dots + 2^na_{2^n}$ be a partial sum of $\sum_{n=1}^{\infty} 2^na_{2^n}$. It is easy to see that,

$$ \begin{aligned} s_n < t_n \end{aligned} $$
$$ \begin{aligned} s_n < t_n \end{aligned} $$

If $\sum_{n=1}^{\infty} 2^na_{2^n}$ converges then, $\lim_{n\rightarrow \infty} t_n = t$. Means that,

$$ \begin{align} s_n < t_n < t \tag{1} \end{align} $$
$$ \begin{align} s_n < t_n < t \tag{1} \end{align} $$

By assumption we know that $a_n$ is monotone decreasing and by $(1)$ it is bounded hence by monotone convergence theorem

$$ \begin{aligned} \lim_{n \rightarrow \infty} s_n = s \end{aligned} $$
$$ \begin{aligned} \lim_{n \rightarrow \infty} s_n = s \end{aligned} $$

for some $s < \infty$, which reads $\sum_{n=1}^{\infty} a_n$ converges.

Now assume, $\sum_{n=1}^{\infty} 2^na_{2^n}$ diverges. See that,

$$ \begin{aligned} 0 &< a_1 + 2a_1 + 4a_2 + \dots + 2^na_{2^n} \\ &< 2(a_1/2 + a_2 + 2a_4 + \dots + 2^{n-1}a_{2^n}) \end{aligned} $$
$$ \begin{aligned} 0 < t_n &= a_1 + 2a_1 + 4a_2 + \dots + 2^na_{2^n} \\ &= 2(a_1/2 + a_2 + 2a_4 + \dots + 2^{n-1}a_{2^n})\\ &< 2(a_1 + a_2 + (a_4 + a_4) + \dots + (a_{2^n} + \dots + a_{2^n}) + \dots + a_n) \\ &< 2(a_1 + a_2 + a_3 + \dots + a_n)\\ &= 2s_n \end{aligned} $$

Therefore, by direct comparison test $\sum_{n=1}^{\infty} a_n$ diverges.

In the next section we will talk about Absolute Convergence, Alternating Series, Rearrangement of Series, Conditional and Absolute Convergence. Bye!