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i have the set $A= \{(1+\frac{1}{n})^{n} : n \in \mathbb{N}^* \}$ and the exercise ask me to find at least $2$ upper bound least equal to $\frac{14}{5}$.

My first question is, how i know $\frac{14}{5}$ is a upper bound of $A$?

My attempt is show doesn't exist $m \in \mathbb{N}^*$ such that $(1+\frac{1}{m})^{m} = \frac{14}{5}$, but i'm stuck, for the other upper bound numbers, i could reason the same way?

Thanks for suggestions

Davis M
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    Hint: what can you say about $\lim_{n\to\infty}(1+\frac{1}{n})^n$? This limit should look familiar. – morrowmh Jan 18 '21 at 01:43
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    Also note $14/5=2.8$. Finally, check that the sequence $a_n:=(1+\frac{1}{n})^n$ is increasing, so it is bounded above by its limit. – morrowmh Jan 18 '21 at 01:52
  • @Michael Morrow the intended way was probably to bound the series expansion directly in some way. (Now addressing the OP) You could use the fact that $n!$ grows faster than $k^n$, and use this to bound small terms by a geometric series. – Benjamin Wang Jan 18 '21 at 02:01
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    An upper bound of $3$ can be found very easily, see this answer. $14/5$ will probably be more difficult. – K.defaoite Jan 18 '21 at 03:33

3 Answers3

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If you are not allowed to use the definition of $e$, then by observing that the sequence terms are bound by (refer to this) $$ \begin{align}\left(1+\frac 1n\right)^n&=\sum_{k=0}^{n}\binom{n}{k}\left(\frac 1n\right)^k\\&=\sum_{k=0}^{n}\frac{n(n-1)\cdots(n-k+1)}{k!}\left(\frac 1n\right)^k\\&=\sum_{k=0}^{n}\frac{1}{k!}\left(1-\frac 1n\right)\left(1-\frac 2n\right)\cdots\left(1-\frac{k-1}{n}\right)\\&\lt \sum_{k=0}^{n}\frac{1}{k!}\\ &=1+\frac{1}{1}+\frac{1}{2}+\frac{1}{3\cdot 2}+\frac{1}{4\cdot 3\cdot 2}+\cdots+\frac{1}{n\cdot(n-1)\cdots 2}\\ &\lt 1+\frac{1}{2^0}+\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+\cdots+\frac{1}{2^{n-1}}\\&=3-\frac{1}{2^{n-1}}\\&\lt 3\end{align} $$ Now to get a stricter bound, since $$ \frac{1}{2^1} + \frac{1}{3\cdot 2}+\frac{1}{4\cdot 3\cdot 2}+\cdots+\frac{1}{n\cdot(n-1)\cdots 2} < \frac {1}{2} \left( 1+\frac{1}{2^0}+\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+\cdots+\frac{1}{3^{n-1}} \right) $$ we have $$ \left(1+\frac 1n\right)^n < 1 + \frac{1}{2^0} + \frac {1}{2} \left(\frac{1}{3^0}+\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+\cdots+\frac{1}{3^{n-1}} \right) < 2 + \frac{1}{2} \frac{1}{1-\frac{1}{3}} = \frac{11}{4} < \frac{14}{5} $$ as required.

macton
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First you should know that: $$ (1) \ln(1+x)=x-\frac{x^2}{2}+O(x^3)$$ Also consider that ln is an injective function meaning if $\ln(A)=1$ then $A=e$
So: $$ \lim_{n\to\infty} \ln(A) = \lim_{n\to\infty} \ln((1+\frac{1}{n})^n = \lim_{n\to\infty} n\ln(1+\frac{1}{n}) $$ Applying (1): $$ \lim_{n\to\infty} n\ln(1+\frac{1}{n}) = \lim_{n\to\infty} n(\frac{1}{n}-\frac{1}{2n^2}+O(\frac{1}{n^3}))$$ $$\lim_{n\to\infty} n(\frac{1}{n}-\frac{1}{2n^2}+O(\frac{1}{n^3})) = \lim_{n\to\infty} (1-\frac{1}{2n}+nO(\frac{1}{n^3})) = 1 $$ Since $$ \lim_{n\to\infty} \ln(A) = 1 $$ Than $$ \lim_{n\to\infty} A = e $$ So e is the least upper bound and any number greater than e is an upper bound (Note: $\frac{14}{5} > e $)

Khosro Moeini
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Actually, the upper bound of this sequence is the standard definition of $e$. More formally, $$ e=\lim_{n\to\infty} (1+{1\over n})^n\approx 2.71828182846 $$ To reject $14\over 5$ as an upper bound, note that $e< 14/5$.

Mostafa Ayaz
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