Are there good lower bounds for the partial sums of the series $\sum 1/\log(n)$?

Question

Consider the partial sums $$S_n = \sum_{k=2}^n\frac{1}{\log(k)}.$$ Are there good lower bounds for $S_n$ as $n\to\infty$ ?

I am not necessarily looking for sharp bounds (although they would be nice), just one that is fair.

Of course a "good" lower bound would be one that go "very" fast to infinity.

Answer 1

Because $1/\ln x$ is decreasing on $[2,\infty),$ we have

$$\tag 1 I_n=\int_2^n \frac{dt}{\ln t} \le S_n.$$

Integrating by parts shows

$$\tag 2 I_n =\frac{n}{\ln n}-\frac{2}{\ln 2}+ \int_2^n \frac{dt}{(\ln t)^2}.$$

Is $(2)$ a decent lower bound of $S_n?$ I would say yes, given that as $n\to \infty,$

$$\tag 3 S_n\sim \frac{n}{\ln n}$$

and

$$\tag 4\int_2^n \frac{dt}{(\ln t)^2} \sim \frac{n}{(\ln n)^2}.$$

You can verify $(3)$ using Stolz-Cesaro, and $(4)$ by using L'Hopital.

Answer 2

We can use $\sqrt{n} \ge \log(n)$ to get a lower bound as

$$\sum_{k=2}^{n} \frac{1}{\sqrt{n}} \approx 2 \sqrt{n} - \text{small constant}$$

Which is a fast enough divergence to $\infty$ I suppose.

Similarly, I believe you should be able to get $n^{1-\epsilon}$ for any $1 > \epsilon > 0$

Answer 3

Note that \begin{align} \frac{1}{\ln k} &= \frac{1}{\ln n - (\ln n - \ln k)}\\ & = \frac{1}{\ln n}\ \frac{1}{1 - \frac{\ln n - \ln k}{\ln n}}\\ &= \frac{1}{\ln n } \Big[1 + \Big(\frac{\ln n - \ln k}{\ln n}\Big) + \Big(\frac{\ln n - \ln k}{\ln n}\Big)^2 + \Big(\frac{\ln n - \ln k}{\ln n}\Big)^3 + \cdots \Big]. \end{align} Let $$\phi_m(n) = \sum_{k=2}^n \frac{1}{\ln n } \Big(\frac{\ln n - \ln k}{\ln n}\Big)^m, \quad m = 0, 1, 2, \cdots$$ We have \begin{align} \phi_0(n) &= \frac{n-1}{\ln n},\\ \phi_1(n) &= \frac{n-1}{\ln n} - \frac{\ln n!}{\ln^2 n}\ \sim\ \frac{n}{\ln^2 n}\quad \mathrm{as}\quad n\to \infty\quad \mathrm{Stirling's}\ \mathrm{formula}\\ \phi_m(n) &\sim \frac{m! n}{\ln^{m+1} n}\quad \mathrm{as}\quad n\to \infty, \quad \mathrm{for}\quad m=2, 3, \cdots\quad \mathrm{Maple}\ \mathrm{software}. \end{align} Thus, we have $\phi_{m+1}(n) = o(\phi_m(n))$ as $n\to \infty$. Remark: We only verified some small $m$ by using Maple software which is not a complete proof.

We have an expansion $$S_n = \phi_0(n) + \phi_1(n) + \cdots,$$ any partial sum of which is a lower bound of $S_n$, for example, $$S_n > \phi_0(n) + \phi_1(n) = \frac{2(n-1)}{\ln n} - \frac{\ln n!}{\ln^2 n}$$ for all $n\ge 2$ (the bound is good for $n\to \infty$).

Remark: Can we show that $$\lim_{n\to \infty} \frac{S_n}{\phi_0(n) + \phi_1(n)} = 1?$$