For $n >0$ ,

Prove that $$ \frac{1}{1+n^2} < \ln(1+ \frac{1}{n} ) < \frac {1}{\sqrt{n}}$$

I really have no clue. I tried by working on $ n^2 + 1 > n > \sqrt{n} $ but it gives nothing.

Any idea?


Note that

$$\frac{1}{1+n^2} < \ln\left(1+ \frac{1}{n} \right)< \frac {1}{\sqrt{n}}\iff\frac{n^2}{1+n^2} < n\ln\left(1+ \frac{1}{n} \right)^n < \frac {n^2}{\sqrt{n}}=n\sqrt n$$

which is true, indeed for $n=1$ the given inequality is true and for $n>1$

$$\frac{n^2}{1+n^2} <1< n\ln\left(1+ \frac{1}{n} \right)^n\leq n\log e=n< n\sqrt n$$

  • 1
    $\begingroup$ nice ;) thank you, that's what i searched $\endgroup$ – Marine Galantin Jan 14 '18 at 12:41

This inequality is correct only for $n >1$ and not for $n >0$ as asked by the person.

For $n >1$, we know that $$\tag1 \frac{1}{n}<\frac{1}{\sqrt n}$$ and $$\tag2 \frac{1}{(1+n^2)} < \frac{1}{(1+n)}$$

Also, $\tag3\frac1{n+1}<\ln\left(1+\tfrac1n\right)<\frac1n, \forall n>0$

So the overall inequality $$\tag4\frac{1}{(1+n^2)} < \frac{1}{(1+n)}<\ln{\left(1+\frac1n\right)}<\frac{1}{n}<\frac{1}{\sqrt n}$$ will be valid only for $n>1$ and not for $n>0$.


I found that with $$ ln(1+x) < x $$ $\forall x\in ]0,+\infty[$

we can easily prove the RLH side of inequality, because $sqrt(x) < x$ still have the problem of the first one.

oh I found something else :

we have also $$ \frac{x}{(1+x)} < ln(1+x) $$ $\forall x\in ]0,+\infty[$

thus here we have $$ ln(1+1/n) >\frac{1}{(1+n)} > \frac{1}{(1+n^2)} $$

I think I have answered my question. thank you :)

  • 1
    $\begingroup$ Use this link as a reference to the inequality you found $\endgroup$ – rtybase Jan 14 '18 at 11:20

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