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Please can you help me with the following question

$$E(x) = \left\{ \left(1 + \frac{x}{n}\right)^n : n \in \mathbb N \right\}$$

Let a(x) = sup E(x) (least upper bound) without finding the sup of E(x)

prove that a(x) < a(y) if 0 < x < y and a(x)a(y) <= {a[ (x+y) / 2] } ^ 2 and a(x + y) = a(x).a(y)

thank you

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Note that for $n\gt|x|$, $$ \left(1+\frac xn\right)^n\left(1-\frac xn\right)^n =\left(1-\frac{x^2}{n^2}\right)^n \le1\tag{1} $$ implies that $$ \left(1+\frac xn\right)^n\le\frac1{\left(1-\frac xn\right)^n}\tag{2} $$ Bernoulli's Inequality, proven at the end of this answer and generalized at the end of this answer, says that $\left(1-\frac xn\right)^n$ is an increasing sequence for $n\gt|x|$. Thus, the right hand side of $(2)$ is a decreasing, positive sequence for $n\gt|x|$, and therefore, bounded.

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Note that $$\binom{n}{k}\frac{x^k}{n^k}=\frac{x^k}{k!}\frac{n(n-1)\cdots(n-k+1)}{n^k}\leq\frac{x^k}{k!}$$ Thus for each $n$, $(1+x/n)^n\leq\sum_{k=0}^n\frac{x^k}{k!}\leq e^x$. Thus the set $E(X)$ is bounded above by $e^x$.

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    $\begingroup$ thanks for the answer, but is e^x the least upper bound of E(x) if yes then how. $\endgroup$ – Charlie Sep 14 '13 at 10:01
  • $\begingroup$ If not then there should be a number, say $y$, such that $y<e^x$ and $y$ is an upper bound for your set $E(X)$. However the terms in $E(X)$ converges to $e^x$, so for every $\epsilon>0$, there is an $N\in\mathbb{N}$, such that for all $n\geq N$, $e^x-(1+x/n)^n<\epsilon$. However take $\epsilon<e^x-y$. The we get a contradiction. $\endgroup$ – Abishanka Saha Sep 14 '13 at 10:49
  • $\begingroup$ thank you could you please hep me with the other part of the question Let a(x) = sup E(x) (least upper bound) without finding the sup of E(x) prove that a(x) < a(y) if 0 < x < y and a(x)a(y) <= {a[ (x+y) / 2] } ^ 2 and a(x + y) = a(x).a(y) $\endgroup$ – Charlie Sep 15 '13 at 7:03

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