# Estimating the sum $\sum_{k=2}^{\infty} \frac{1}{k \ln^2(k)}$

By integral test, it is easy to see that $$\sum_{k=2}^{\infty} \frac{1}{k \ln^2(k)}$$ converges. [Here $\ln(x)$ denotes the natural logarithm, and $\ln^2(x)$ stands for $(\ln(x))^2$]

I am interested in proving the following inequality (preferrably using integral calculus) $$\sum_{k=2}^{\infty} \frac{1}{k \ln^2(k)}>2$$

By wolfram alpha, the actual value of the sum is about 2.10974. Since $\frac{1}{k \ln^2(k)}$ is decreasing, we have
$$\sum_{k=2}^{\infty} \frac{1}{k \ln^2(k)}\ge \int_2^{\infty} \frac{1}{x \ln^2(x)} dx=\frac{1}{\ln(2)}\approx 1.4427$$ So this lower bounded is weaker than desired.

My motivation for asking this question is that by being able to estimate this particular sum will hopefully teach me a general technique which I may try applying to sums of the form $\sum_{k=1}^{\infty} f(k)$.

Thanks!

• As a quick thought, I recommend shifting the index down one and estimating $\ln(1+x)$ as a polynomial using its Maclaurin series. – Jon Claus May 19 '13 at 17:58
• I'm afraid there's no closed form solution to this. – Inceptio May 19 '13 at 18:08
• @Inceptio I don't think that the OP is wanting the precise (closed form) sum, but would be happy enough to prove the inequality. Is that correct, Prism? – Namaste May 19 '13 at 18:24
• Numerical checking gives $$\frac{1}{2 \log^2(2)}+\frac{1}{3 \log ^2(3)}+\int_4^{\infty } \frac{1}{x \log ^2(x)} \, dx=2.03821$$ – Andrew May 19 '13 at 18:33
• @amWhy: Yes, that is correct :) – Prism May 19 '13 at 18:36

\begin{align*} \sum_{k=2}^{\infty} \frac1{k\ln^2k} &= \frac1{2\ln^22} + \frac1{3\ln^23} + \sum_{k=4}^{\infty} \frac1{k\ln^2k} \\ &\ge \frac1{2\ln^22} + \frac1{3\ln^23} + \int_4^{\infty} \frac{dx}{x\ln^2x} \\ &= \frac1{2\ln^22} + \frac1{3\ln^23} + \frac{1}{\ln4} > 2.038. \end{align*} (I see Andrew just wrote this in a comment.)

• I like this technique! Very simple and to the point. – Prism May 19 '13 at 18:38
• I had a tough time getting your answer, till I got back to the question to find log(x) denotes natural logarithm (+1). – Inceptio May 19 '13 at 18:41
• I think that we may not use such estimates for $\log 2, \log 3, \log 4$. It is obvious that from some $k\in \mathbb N$ our sum will be so small that we will have reached our target of $2$ with the terms out of the sum. Luckily here it happens with the first two terms. I mean, I would be satisfied if for example we reached a result involving less estimates. – Dimitris May 19 '13 at 18:46
• How does one prove that last result though. It can be done numerically easily, but so could the original problem. – Jon Claus May 19 '13 at 20:48
• There are lots of ways of approximating $\ln x$ arbitrarily closely: via its power series, for example, or with a Riemann sum approximating the appropriate integral. So getting upper bounds on $\ln 2$ and $\ln 3$ that are within 1% of the true values is not hard, and that will establish that the last expression exceeds $2$. – Greg Martin May 19 '13 at 23:03

I know the question is to show that the sum is greater than $2$, and Greg Martin's answer does that perfectly; however, I thought it might be interesting to compute the actual value of the sum.

Using the Euler-Maclaurin Sum Formula, we get \begin{align} &\sum_{k=2}^n\frac1{k\log(k)^2}\\ &=C-\frac1{\log(n)}+\frac1{2n\log(n)^2}-\frac1{12n^2}\left(\frac1{\log(n)^2}+\frac2{\log(n)^3}\right)\\ &+\frac1{360n^4}\left(\frac3{\log(n)^2}+\frac{11}{\log(n)^3}+\frac{18}{\log(n)^4}+\frac{12}{\log(n)^5}\right)\\ &\scriptsize\,-\,\frac1{15120n^6}\left(\frac{60}{\log(n)^2}+\frac{274}{\log(n)^3}+\frac{675}{\log(n)^4}+\frac{1020}{\log(n)^5}+\frac{900}{\log(n)^6}+\frac{360}{\log(n)^7}\right)\\ &\tiny\,+\,\,\frac1{50400n^8}\left(\frac{210}{\log(n)^2}+\frac{1089}{\log(n)^3}+\frac{3283}{\log(n)^4}+\frac{6769}{\log(n)^5}+\frac{9800}{\log(n)^6}+\frac{9660}{\log(n)^7}+\frac{5880}{\log(n)^8}+\frac{1680}{\log(n)^9}\right)\\ &\tiny\,-\,\,\frac1{1995840n^{10}}\left(\frac{15120}{\log(n)^2}+\frac{85548}{\log(n)^3}+\frac{293175}{\log(n)^4}+\frac{723680}{\log(n)^5}+\frac{1346625}{\log(n)^6}+\frac{1898190}{\log(n)^7}+\frac{1984500}{\log(n)^8}+\frac{1461600}{\log(n)^9}+\frac{680400}{\log(n)^{10}}+\frac{151200}{\log(n)^{11}}\right)\\ &+O\!\left(\frac1{n^{12}\log(n)^2}\right)\tag1 \end{align} Therefore, plugging $n=1000$ into $(1)$, we get \begin{align} \sum_{k=2}^\infty\frac1{k\log(k)^2} &=C\\ &=2.109742801236891974479257197616551326\tag2 \end{align}

• @Waiting: Congratulations! – robjohn Jun 13 '18 at 14:18
• robjohn, thank you!^^^ – user 1357113 Jun 13 '18 at 15:53