Htykuut

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 $log_2 x > sumlimits_{i le x}frac{1}{i}$

(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) $(x+1)log_2left(frac{x+1}{x}right) > 1$

(5) $log_2(x+1) - log_2(x) > frac{1}{x+1}$

(6) $log_2(x+1) - log_2(x) + log_2(x) = log_2(x+1) > sumlimits_{i le x+1}frac{1}{i}$

We have that

$$1 + frac{1}{2} + frac{1}{3} + dots + frac{1}{x} < log_2 x iff 1 + frac{1}{2} + frac{1}{3} + dots + frac{1}{x} < frac{ln x}{ln 2}$$

$$ln 2left(1 + frac{1}{2} + frac{1}{3} + dots + frac{1}{x} right)< ln x$$

then recall that by harmonic series

$$sum_{k=1}^x frac1k simln x+gamma+frac1{2x}$$

then

$$ln 2left(1 + frac{1}{2} + frac{1}{3} + dots + frac{1}{x} right) sim ln 2ln x+gammaln 2+frac{ln 2}{2x}stackrel{?}<ln x$$

$$(1-ln 2)ln x> gammaln 2+frac{ln 2}{2x}$$

which is true for $x$ sufficiently large.

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
$$log_2(n)+frac{1}{n+1} < log_2(n+1)$$
$$frac{1}{n+1}< log_2left(1+frac{1}{n}right)$$
$$2^{frac{1}{n+1}}<1+frac{1}{n}$$
$$2<left(1+frac{1}{n}right)^{n+1}$$
$$2<left(1+frac{1}{n}right) left(1+frac{1}{n}right)^n$$
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 $forall ngeq 6 land n in mathbb{N}$ (it's true for $n=5$ as well, but Bernoulli is not helping in that case).

Yes because $H_n-log n$ is bounded by $1$ and $log_2n$ is a multiple of $log n$.