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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.

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  • $\begingroup$ Checked with Python. Inequality holds: left side is about $6.9×10^7$. $\endgroup$ – Parcly Taxel Oct 3 '16 at 1:58
  • $\begingroup$ @ParclyTaxel Thanks, that's what Wolfram was getting at. But because the left hand side is $7\times 10^7$, and therefore very close, is there any way to ensure that approximation error isn't producing a smaller than actual value output? This inequality arises in a paper I am writing so I don't want things to seem "hand wavy" if you will. $\endgroup$ – Will Fisher Oct 3 '16 at 2:02
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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: $$ \frac{1.25506(2\cdot10^9-1)}{\log(2\cdot10^9-1)}-\frac{10^9}{\log10^9}<\frac{2.51012 \cdot 10^9}{\log 10^9}-\frac{10^9}{\log 10^9}<\frac{1.68}{\log 10}\cdot 10^8\approx 0.73\cdot 10^8. $$

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