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$$\frac{1}{n+1}< \log(1+ 1/n)$$

Any ideas? I tried estimating the difference between $1/n$ and the logarithm and comparing with $1/n-1/(n+1)$ but I miserably failed.

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Did you try considering the function $f(x)=\log(1+x^{-1})-(x+1)^{-1}$? – egreg Apr 5 '14 at 16:38
Well I suppose I could do that, but this was an exercise after a chapter on successions and series, so I don't think it would be fair. – Gennaro Marco Devincenzis Apr 5 '14 at 16:39
You can write the right hand side as $\log \frac{n+1}{n}$, maybe that gives you an idea. – Daniel Fischer Apr 5 '14 at 16:42
Taylor series?${}{}{}{}{}{}$ – Sanath K. Devalapurkar Apr 5 '14 at 16:45
Try integrating 1/x over a suitable range. Maybe still not fair though! – Paul Apr 5 '14 at 16:47
up vote 4 down vote accepted

The inequality is equivalent to $$ (n+1)\log\left(1+\frac{1}{n}\right)>1 $$ or $$ \left(1+\frac{1}{n}\right)^{n+1}>e $$ which is known to be true: it's usually part of the definition of $e$ by means of sequences.

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makes sense. thank you. – Gennaro Marco Devincenzis Apr 5 '14 at 16:49
Could you please give more explanation. I know that the definition is usually $$\lim_{n \to \infty}\left(1+\frac{1}{n}\right)^{\!n} = \mathrm{e}$$ – Fly by Night Apr 5 '14 at 16:52
@FlybyNight If you change the exponent to $n+1$, you get a decreasing sequence converging to $e$, and $(1+1/n)^n<e<(1+1/n)^{n+1}$ for all $n$. – egreg Apr 5 '14 at 16:53
Ah, of course. Thank you. – Fly by Night Apr 5 '14 at 16:56

Using the integral definition of the logarithm you have, for $n>0$, $$ \log(1+\frac1n)=\int_1^{1+\frac1n}\,\frac{1}x\,dx>\int_1^{1+\frac1n}\frac1{1+\frac1n}\,dx=\frac{\frac1n}{1+\frac1n}=\frac1{n+1} $$ (note that the inequality is strict because $1/x$ is monotone and non-constant).

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You need to disprove equality. – Fly by Night Apr 5 '14 at 16:58
The inequality is strict because $1/x$ is monotone and non-constant. – Martin Argerami Apr 5 '14 at 17:13
Thanks for making the improvement. A very nice answer. – Fly by Night Apr 5 '14 at 17:22
Thanks for noticing. I'm editing the answer to include mention of that. – Martin Argerami Apr 5 '14 at 18:43

We apply the mean value theorem to the function $\log$ on the interval $\left(1,1+\frac1n\right)$: $$\log\left(1+\frac1n\right)=\frac1n\frac1c$$ where $$1<c<1+\frac1n\Rightarrow \frac1c>\frac n{n+1}$$ so we conclude that $$\log\left(1+\frac1n\right)>\frac 1{n+1}$$

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You can also use Mean Value Theorem over an interval [1,x] of log(x)

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