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Let $A=\left\{ \left(\dfrac{m+n+1}{m+n}\right)^{m+n};\ m,n\in\mathbb{N}^* \right\}$.

Calculate the supremum of $A$ ($\sup A$).

I tired

To find the first few elements of $A$

$\left(\dfrac{3}{2}\right)^{2}$ It appears that $\sup A=\left(\dfrac{3}{2}\right)^{2}$ We use the definitions of $\sup A$ to prove this result

For any $x\in A$ we have $x=\left(\dfrac{m+n+1}{m+n}\right)^{m+n}$

any help would be appreciated

Community
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Educ
  • 4,780

2 Answers2

2

Hint

$$\lim_{n\to\infty}\left(1+\frac1n\right)^n=e$$

EDIT:

$$\left(\frac{m+n+1}{m+n}\right)^{m+n}=\left(1+\frac{1}{m+n}\right)^{m+n}$$

Since $m+n$ can take any natural value $\ge 2$, and the sequence $$\left(1+\frac 1n\right)^n$$ is increasing and bounded, the supremum of $A$ is the limit of the sequence.

ajotatxe
  • 65,084
2

Remember that the sequence $$ \left(1+\frac{1}{k}\right)^k,\qquad k\in\mathbb N, $$ for $k\in\mathbb N$ is monotonically increasing and tends to $e$ as $k\to\infty$, so we guess that $\sup(A)=e$. To see this, fix $m,n\in\mathbb N$. Then for $k=m+n$ we have due to the monotony that $$ \left(\frac{m+n+1}{m+n}\right)^{m+n}=\left(1+\frac{1}{k}\right)^k\leq e, $$ and since $m$ and $n$ were arbitrary, we have $x\leq e$ for each $x\in A$, so $e$ is an upper bound of $A$. But it must be the smallest one, since for $m=0$ and arbitrary $n$ the numbers $$ \left(\frac{m+n+1}{m+n}\right)^{m+n}=\left(1+\frac{1}{n}\right)^n $$ form a sequence of elements of $A$ which tends to $e$. (If you are not allowed to choose $m=0$, then you can also take $m=n$ (or even $m=1$). Then you end up with $$ \left(\frac{m+n+1}{m+n}\right)^{m+n}=\left(1+\frac{1}{2n}\right)^{2n}, $$ and you can use the same argument as before (note that $(1+\frac{1}{2n})^{2n}$ is a subsequence of $(1+\frac{1}{n})^n$).

sranthrop
  • 8,497
  • could you see this http://math.stackexchange.com/questions/1027554/series-and-inequality – Educ Nov 23 '14 at 15:27
  • to show it that's the smallest one it's suffice to take sequence of elements of A and prove it that tends to e do you have any idea how to prove $\left(1+\frac{1}{k}\right)^k$, increasing and tends to $e$ – Educ Dec 03 '14 at 20:18
  • this is very well-known and can easyly be found everywhere in the web. For example: http://math.stackexchange.com/questions/64860/proving-bigl1-frac1n1-bigrn1-gt-1-frac1nn – sranthrop Dec 03 '14 at 20:45