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TL;DR Is there any way to solve this by comparison?

Context: This question is posed in my text book after the integral test, the comparison test, and the limit-comparison test, but before any of the other tests.

I'm pretty sure the answer is $r>1$, and I can arrive at that with the integral test. But that requires some painful integration, and I can't help but think that I should be able to compare the sum with $\sum\frac{1}{n^r}$, which I know converges when $r>1$.

Since $\ln(n)<n$, I tried comparing with $\sum\frac{n}{n^r}=\sum\frac{1}{n^{r-1}}$ to get $r>2$. And while it certainly does converge when $r>2$, that's not entirely correct.

Can this be done by comparison?

Asaf Karagila
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Chris Barrick
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4 Answers4

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By Comparison, our series diverges if $r\le 1$. We show it converges if $r\gt 1$. Let $r=1+\delta$. Then $n^r=n^{\delta/2}n^{1+\delta/2}$.

Note that $\displaystyle\lim_{n\to\infty}\dfrac{\ln n}{n^{\delta/2}}=0$. So by Limit Comparison with $\displaystyle\sum_{1}^\infty \dfrac{1}{n^{1+\delta/2}}$, our series converges.

Remark: We sliced off a small piece $\delta/2$ of $r$. The term $n^{\delta/2}$ in the long run grows fast enough to crush $\ln n$. But there enough of $r$ left to force convergence.

André Nicolas
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You can use both the comparison test as shown by André Nicolas or the integral test, indeed:

First if $r\leq0$ the series is divergent since the general term of this series doesn't converge to $0$.

Now let $r>0$, so by derivative we can see easly that the positive function $x\to\frac{\ln x}{x^r}$ is decreasing on some interval $[a,+\infty)$, moreover by integration by parts we have $$\int_a^n\frac{\ln x}{x^r}dx=\frac{1}{1-r}\left[x^{1-r}\ln x\right]_a^n-\frac{1}{1-r}\int_a^n\frac{dx}{x^{r}}$$ hence it's clear the improper integral $$\int_a^\infty\frac{\ln x}{x^r}dx$$ is convergent if and only if $r>1$.

Daniel Fischer
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Let $g(s) = \displaystyle \sum_{k=1}^{\infty} \dfrac1{k^s}$ and $f_{n}(s) = \displaystyle \sum_{k=1}^n \dfrac{\log(k)}{k^s}$. Clearly, $f_n(s)$ is an increasing function of $n$. We then have $$f_{2n+1}(s) = \sum_{k=1}^{2n+1} \dfrac{\log(k)}{k^s} = \sum_{k=2}^{2n+1} \dfrac{\log(k)}{k^s} = \sum_{k=1}^{n} \dfrac{\log(2k)}{(2k)^s} + \sum_{k=1}^{n} \dfrac{\log(2k+1)}{(2k+1)^s} \tag{$\star$}$$ Now $\log(x)/x$ is a decreasing function for $x \geq 3$. Hence, we get $$\sum_{k=1}^{n} \dfrac{\log(2k+1)}{(2k+1)^s} < \dfrac{\log(3)}{3^s} + \sum_{k=2}^{n} \dfrac{\log(2k)}{(2k)^s} = \underbrace{\dfrac{\log(3)}{3^s} -\dfrac{\log(2)}{2^s}}_{c(s)} + \sum_{k=1}^{n} \dfrac{\log(2k)}{(2k)^s} \tag{$\perp$}$$ Plugging $(\perp)$ in $(\star)$, we get that \begin{align} f_{2n+1}(s) & < c(s) + 2 \sum_{k=1}^{n} \dfrac{\log(2k)}{(2k)^s} = c(s) + 2^{1-s} \cdot \sum_{k=1}^{n} \dfrac{\log(2k)}{k^s}\\ & = c(s) + 2^{1-s} \log(2) \cdot \underbrace{\sum_{k=1}^{n} \dfrac1{k^s}}_{< g(s)} + 2^{1-s} \cdot \underbrace{\sum_{k=1}^{n} \dfrac{\log(k)}{k^s}}_{f_n(s) < f_{2n+1}(s)}\\ & < c(s) + 2^{1-s} \log(2) g(s) + 2^{1-s} f_{2n+1}(s) \end{align} This gives us $$\left(1-2^{1-s}\right)f_{2n+1}(s) < c(s) + 2^{1-s} \log(2) g(s) \implies f_{2n+1}(s) < \dfrac{c(s) + 2^{1-s} \log(2) g(s)}{1-2^{1-s}}$$ And clearly, the right hand side is finite for $s>1$. Hence, we have $$f_{2n}(s) < f_{2n+1}(s) < \dfrac{c(s) + 2^{1-s} \log(2) g(s)}{1-2^{1-s}} < \infty$$ Hence, $\lim_{n \to \infty} f_n(s)$ exists as a real number for $s>1$.

This is inspired by Joriki's proof here.

Community
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It can be shown that for all $\varepsilon > 0$ there exist $x_0\in \mathbb{R}$ such that for all $x>x_0$ the inequality $$\ln x <x^{\varepsilon} $$ holds (by proving $\lim_{x \to \infty} \frac{\ln x}{x^\varepsilon} = 0$ applying L'hopital's rule).

Combining this fact with the comparison test (noticing that the terms are positive) will yield the requested result.

Amihai Zivan
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