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Given the inequality $\ln(1+x)\leq x$. Is it possible to prove that $\log_2(1+\frac{1}{x}) \geq \frac{1}{x}$ for $x \geq 1$?

When $x\to \infty$ we have $1/x \to 0$, so we can use the Maclaurin expansion of $\ln(1+\frac{1}{x})=\frac{1}{x}-\frac{1}{2x^2}+\frac{1}{3x^3}-...$ to see that $\ln(1+\frac{1}{x})\leq \frac{1}{x}$, since the sum is negative if we exclude the first term for any $x\geq1$. However, with $\log_2(1+\frac{1}{x})$ that's not the case.

Can anyone please explain?

giorgioh
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    The Maclaurin expansion is in powers of $x$ and not $1/x$, see here. – Gary Jan 29 '22 at 12:37
  • @Gary Isn't it possible to substitute $x$ for $1/x$? I know there needs to be an argument regarding uniform convergence of the series but I didn't dig very deep into it. – giorgioh Jan 29 '22 at 12:50
  • But you wrote $\ln(1+x)$ and not $\ln(1+1/x)$. Please revise your question and fix the typos. – Gary Jan 29 '22 at 12:51
  • My bad, I'll correct this. – giorgioh Jan 29 '22 at 12:51
  • Show that $\frac{{\ln (1 + 1/x)}}{{1/x}}$ is monotonically increasing for $x\geq 1$. This means that $\frac{{\ln (1 + 1/x)}}{{1/x}}\ge\frac{{\ln (1 + 1/1)}}{{1/1}}=\ln 2$, i.e., $\ln(1+1/x)/\ln 2\geq 1/x$, i.e., $\log_2(1+1/x) \geq 1/x$. – Gary Jan 29 '22 at 13:02

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