I was wondering if anyone could help me show that $$\frac{1.25506(2\cdot 10^9-1)}{\ln(2\cdot 10^9 - 1)}-\frac{10^9}{\ln(10^9)}\le \frac{3}{4}\cdot 10^8$$ These values come very close to being equal, and results from Wolfram|Alpha vary depending on how you ask it the problem. My gut says that this inequality does hold but I haven't been able to prove it because even the slightest of approximations can throw it off. Can anyone help me with tackling this mess? Note that I may be wrong on the inequality holding, so any sort of resolution would be great.
2026-04-02 04:27:28.1775104048
Inequality of Large, Non-Integer Numbers
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They're not that close.... the left-hand side and the right-hand side differ by almost 10%. Very rough bounds on the left-hand side get you where you need to go: $$ \frac{1.25506(2\cdot10^9-1)}{\log(2\cdot10^9-1)}-\frac{10^9}{\log10^9}<\frac{2.51012 \cdot 10^9}{\log 10^9}-\frac{10^9}{\log 10^9}<\frac{1.68}{\log 10}\cdot 10^8\approx 0.73\cdot 10^8. $$