Raabe's Test

State Raabe's test and prove it by comparison with the $p$ -series.

L=\lim_{k\to\infty}k\left(\frac{a_k}{a_{k+1}}-1\right),\qquad L>1\Rightarrow\sum a_k\ \text{converges},\quad L<1\Rightarrow\sum a_k\ \text{diverges}

n = 100

:

R_{100} \approx 1.4975

— settling above the boundary

1

, climbing toward

\frac{3}{2}

.

Show proof

Raabe's test: if $a_k > 0$ and $L = \lim_{k\to\infty} k\left(\frac{a_k}{a_{k+1}} - 1\right)$ exists, then $\sum a_k$ converges when $L > 1$ and diverges when $L < 1$ . (Read every inequality below as holding for all large $k$ .)
The comparison benchmark is the $p$ -series $b_k = \frac{1}{k^r}$ . Since $\frac{b_k}{b_{k+1}} = \left(\frac{k+1}{k}\right)^r = \left(1+\frac{1}{k}\right)^r$ and $\left(1+\frac{1}{k}\right)^r = 1+\frac{r}{k}+o\left(\frac{1}{k}\right)$ , we get $k\left(\frac{b_k}{b_{k+1}}-1\right) \to r$ — so $b_k$ is a series whose Raabe value is exactly $r$ .
If $L > 1$ : pick $r$ with $1 < r < L$ . Since $k\left(\frac{a_k}{a_{k+1}}-1\right) \to L$ and $k\left(\frac{b_k}{b_{k+1}}-1\right) \to r$ , eventually $\frac{a_k}{a_{k+1}}-1 > \frac{b_k}{b_{k+1}}-1$ , i.e. $\frac{a_{k+1}}{a_k} < \frac{b_{k+1}}{b_k}$ . Then $a_{k+1}b_k < a_k b_{k+1}$ , i.e. $\frac{a_{k+1}}{b_{k+1}} < \frac{a_k}{b_k}$ , so $\frac{a_k}{b_k}$ decreases and $a_k \le Cb_k$ ; comparing with the convergent $\sum b_k$ ( $r > 1$ ), $\sum a_k$ converges.
If $L < 1$ : picking $r$ with $L < r < 1$ reverses the inequalities to $\frac{a_{k+1}}{a_k} > \frac{b_{k+1}}{b_k}$ ; then $\frac{a_k}{b_k}$ increases, so $a_k \ge cb_k$ ( $c > 0$ ), and comparing with the divergent $\sum b_k$ ( $r < 1$ ) gives divergence of $\sum a_k$ .

With $u_k = \frac{\binom{2k}{k}}{4^k}$ and $a_k = \frac{u_k}{2k+1}$ , prove $\sum_{k=1}^{\infty} a_k$ converges via Raabe's test, and show why the ratio test stays silent.

\frac{a_k}{a_{k+1}}=\frac{2(k+1)(2k+3)}{(2k+1)^2},\qquad R_k=k\left(\frac{a_k}{a_{k+1}}-1\right)=\frac{k(6k+5)}{4k^2+4k+1}\to\frac{3}{2}>1

n = 30

:

R_{30} \approx 1.4915

(above the boundary

1

),

r_{30} \approx 0.9526

—

r_k

clings to 1 while

R_k

settles above 1.

Show proof

First the ratio: $\frac{u_{k+1}}{u_k} = \frac{(2k+2)(2k+1)}{(k+1)^2}\cdot\frac{1}{4} = \frac{2k+1}{2k+2}$ , so $\frac{a_k}{a_{k+1}} = \frac{u_k}{u_{k+1}}\cdot\frac{2k+3}{2k+1} = \frac{2k+2}{2k+1}\cdot\frac{2k+3}{2k+1} = \frac{2(k+1)(2k+3)}{(2k+1)^2}$ .
Expanding, the numerator is $2(k+1)(2k+3) = 4k^2+10k+6$ and the denominator is $(2k+1)^2 = 4k^2+4k+1$ , so $\frac{a_k}{a_{k+1}}-1 = \frac{6k+5}{4k^2+4k+1}$ . Hence $R_k = k\frac{6k+5}{4k^2+4k+1} \to \frac{6}{4} = \frac{3}{2} > 1$ , and $\sum a_k$ converges by Raabe's test.
Meanwhile $\frac{a_{k+1}}{a_k} = \frac{4k^2+4k+1}{4k^2+10k+6} \to 1$ , so the ratio test is inconclusive — the limit is $1$ : starting from $r_1 = \frac{9}{20}, r_2 = \frac{25}{42}, r_3 = \frac{49}{72}$ it crawls upward toward $1$ . Raabe's $R_k$ looks one order deeper than the ratio.

For $a_k = \frac{1}{k(\ln k)^p}$ ( $k \ge 2$ ), show the Raabe limit is $1$ for every $p$ , and prove the boundary $L = 1$ is genuinely inconclusive — $\sum_{k\ge 2}\frac{1}{k(\ln k)^2}$ converges while $\sum_{k\ge 2}\frac{1}{k\ln k}$ diverges — via the integral test.

k\left(\frac{a_k}{a_{k+1}}-1\right)=1+\frac{p}{\ln k}+o(1)\to 1\qquad(\text{for every }p)

n = 200

:

R_{200} \approx 1.3786

for

p = 2

,

R_{200} \approx 1.1892

for

p = 1

— both crawling down to

1

.

Show proof

$\frac{a_k}{a_{k+1}} = \frac{k+1}{k}\left(\frac{\ln(k+1)}{\ln k}\right)^p = \left(1+\frac{1}{k}\right)\left(1+\frac{\ln(1+1/k)}{\ln k}\right)^p$ . With $\ln\left(1+\frac{1}{k}\right) = \frac{1}{k}+O\left(\frac{1}{k^2}\right)$ and $(1+x)^p = 1+px+o(x)$ , the parenthesis is $1+\frac{p}{k\ln k}+o\left(\frac{1}{k}\right)$ .
Multiplying out, $\frac{a_k}{a_{k+1}} = 1+\frac{1}{k}+\frac{p}{k\ln k}+o\left(\frac{1}{k}\right)$ , so $k\left(\frac{a_k}{a_{k+1}}-1\right) = 1+\frac{p}{\ln k}+o(1) \to 1$ — the Raabe limit is $1$ for every $p$ .
Yet the two series have opposite fates. Substituting $t = \ln x$ , $\int_2^{\infty}\frac{dx}{x(\ln x)^p} = \int_{\ln 2}^{\infty}\frac{dt}{t^p}$ ; by the integral test $\sum_{k\ge 2}\frac{1}{k(\ln k)^2}$ converges ( $\int_{\ln 2}^{\infty}t^{-2}\,dt = \frac{1}{\ln 2}$ ) while $\sum_{k\ge 2}\frac{1}{k\ln k}$ diverges ( $\int_{\ln 2}^{\infty}t^{-1}\,dt = \infty$ ). At $L = 1$ Raabe's test is genuinely silent.

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