# Showing $\int_{0}^{1}\frac{\log(1+x+x^{2}+\cdots+x^{2^{k}-1})}{x}\,dx <2-\frac{1}{2^{k-1}}$ [closed]

Let $x\in (0,1)$. Then

$$\int_{0}^{1}\frac{\log(1+x+x^{2}+\cdots+x^{2^{k}-1})}{x}\,dx <2-\frac{1}{2^{k-1}}$$

How does one prove this inequality. It's clear that if $k\to \infty$ the integral evaluates to $\zeta(2)$ which is less than $2$.

## closed as off-topic by Saad, Namaste, Ethan Bolker, Xander Henderson, AweyganJun 19 '18 at 17:56

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• It's clear that the sequence of integrals is increasing. And $\pi^2/6\approx 1.64$ For $k=3$, the r.h.s. is 1.75, so really you'd just need to show this holds for $k=1,2$. – Alex R. Nov 10 '17 at 21:15

$(1+x+x^2+ \cdots +x^{2^k - 1}) = \frac {1-x^{2^k}}{1-x}$

$\ln(1-x) = -\sum_\limits{n=1}^{\infty} \frac {x}{n}\\ \ln(1-x) < -x$

$\frac {\ln(1-x^{2^k}) - \ln (1-x)}{x} < -x^{2^k-1} - \frac {\ln(1-x)}{x}\\ \int_0^1\frac {\ln(1-x^{2^k}) - \ln (1-x)}{x} \ dx< \int_0^1-x^{2^k-1} -\frac {\ln(1-x)}{x}\ dx\\ \int_0^1-x^{2^k-1}\ dx = -\frac {1}{2^k}\\ \int_0^1 -\frac {\ln(1-x)}{x}\ dx = \int\sum_\limits{n=1}^{\infty} \frac {x^{-1}}{n} = \sum_\limits{n=1}^{\infty} \frac {1}{n^2} = \zeta{(2)}$

I get

$\int_0^1 \frac {\ln (1+x + x^2 +\cdots+ x^{2^k-1})}{x} < \zeta (2) - \frac 1{2^k}$

Notice for any $y > 0$, $\log(1+y) < y$. Together with the identity $$1 + x + \cdots + x^{2^k-1} = (1+x)(1+x^2)\cdots(1+x^{2^{k-1}})$$ We find $$\int_0^1 \frac1x\log\sum_{j=0}^{2^k-1}x^j\, dx = \int_0^1 \frac1x\log\prod_{j=0}^{k-1}(1+x^{2^j})\,dx = \int_0^1 \frac1x\sum_{j=0}^{k-1}\log(1+x^{2^j})\,dx\\ < \int_0^1 \sum_{j=0}^{k-1} x^{2^j-1} dx = \sum_{j=0}^{k-1}\frac{1}{2^j} = \frac{1 - \frac{1}{2^k}}{1-\frac12} = 2 - \frac{1}{2^{k-1}}$$

• Very nice. Thanks for the help. – crskhr Nov 11 '17 at 5:44