3
$\begingroup$

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?

$\endgroup$
4
$\begingroup$

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 $n>1$

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

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

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 $n>0$.

$\endgroup$
0
$\begingroup$

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 :)

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.