Raabe's Test

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 $\sum a_k$.

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.

$\frac{a_k}{a_{k+1}} = \frac{k+1}{k}\left(\frac{\ln(k+1)}{\ln k}\right)^p = \left(1+\frac{1}{k}\right)\left(1+\frac{\ln(1+1/k)}{\ln k}\right)^p$. 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.

First verify the integral test's hypotheses: for $x \ge 2$, both $x$ and $(\ln x)^p$ ($p = 1, 2$) are positive and increasing, so their product $x(\ln x)^p$ is increasing, hence the integrand $\frac{1}{x(\ln x)^p}$ is positive and decreasing.

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.