Is $1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{x} < \log_2 x$ If $x \ge 5$, is $1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{x} < \log_2 x$
I believe the answer is yes.
Here is my thinking:
(1)  $\log_2{5} > 2.32 > 2.284 > 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5}$
(2) Assume up to $x$ that $\log_2 x > \sum\limits_{i \le x}\frac{1}{i}$
(3) $(x+1)^{x+1} > {{x+1}\choose{x+1}}x^{x+1} + {{x+1}\choose{x}}x^x = x^{x+1} + (x+1)x^x = 2x^{x+1} + x^x > 2(x^{x+1})$
(4) $(x+1)\log_2\left(\frac{x+1}{x}\right) > 1$
(5) $\log_2(x+1) - \log_2(x) > \frac{1}{x+1}$
(6) $\log_2(x+1) - \log_2(x) + \log_2(x) = \log_2(x+1) > \sum\limits_{i \le x+1}\frac{1}{i}$
 A: We have that
$$1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{x} < \log_2 x \iff 1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{x} < \frac{\ln x}{\ln 2}$$
$$\ln 2\left(1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{x} \right)< \ln x$$
then recall that by harmonic series
$$\sum_{k=1}^x \frac1k \sim\ln x+\gamma+\frac1{2x}$$
then
$$\ln 2\left(1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{x} \right) \sim \ln 2\ln x+\gamma\ln 2+\frac{\ln 2}{2x}\stackrel{?}<\ln x$$
$$(1-\ln 2)\ln x> \gamma\ln 2+\frac{\ln 2}{2x}$$
which is true for $x$ sufficiently large.
A: We can use induction as well. If we assume that it's true for $n$:
$$1+\frac{1}{2}+...+\frac{1}{n}+\frac{1}{n+1}<\log_2(n)+\frac{1}{n+1}$$
So we need to prove that
$$\log_2(n)+\frac{1}{n+1} < \log_2(n+1)$$
$$\frac{1}{n+1}< \log_2\left(1+\frac{1}{n}\right)$$
$$2^{\frac{1}{n+1}}<1+\frac{1}{n}$$
$$2<\left(1+\frac{1}{n}\right)^{n+1}$$
$$2<\left(1+\frac{1}{n}\right) \left(1+\frac{1}{n}\right)^n$$
And it's true, because
$$1+\frac{1}{n}>1$$
And by Bernoulli's inequality:
$$\left(1+\frac{1}{n}\right)^n \geq 1+n \frac{1}{n}=2$$
So we just need to find a good starting $n$. Let's check it for $n=6$:
$$1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}<\log_2(6)$$
$$\frac{9}{20}<\log_2({6/4})$$
$$2^{\frac{9}{20}}<\frac{6}{4}$$
And by Bernoulli's inequality,
$$2^{\frac{9}{20}}<1+\frac{9}{20}<1+\frac{10}{20}=\frac{6}{4}$$
So it's true $\forall n\geq 6 \land n \in \mathbb{N}$ (it's true for $n=5$ as well, but Bernoulli is not helping in that case).
A: Yes because $H_n-\log n$ is bounded by $1$ and $\log_2n$ is a multiple of $\log n$.
