How to verify that $\frac{68}{25}$ is an upper bound for the set $A=\{(1+\frac{1}{n})^n|n\in\Bbb N^*\}$ I've tried to prove it aiming for a contradiction:If $\frac{68}{25}$ is not an upper bound of A, then, $\exists n\in\Bbb N^*$ such that $$\frac{68}{25}\lt(1+\frac{1}{n})^n\Rightarrow\frac{(68)^{1/n}}{(25)^{1/n}}-\frac{1}{n}<1\Rightarrow n(68)^{1/n}-(25)^{1/n}\lt n(25)^{1/n}\Rightarrow n(68)^{1/n}\lt(25)^{1/n}(n+1)$$ At this part i get stuck, because, we see that $n<n+1$ and instinctively i'd say that
$(68)^{1/n}<(25)^{1/n}$, apparently, i´ve found a contradiction: $68<25$. But i've found that in general if
we have $a<b$ and $ac<bd$ it´s not always true that $c<d$, for instance, $2<3$ and $2\sqrt2<3(1)$
don't imply that 
$\sqrt2<1$. What can i do?, Please, help me. THANKS
 A: The expression $\left(1 + \frac{1}{n}\right)^n$ is increasing, and its limit as $n \to \infty$ is $$e = \sum_{k = 0}^{\infty} \frac{1}{k!} ,$$
so as hinted in the comments, the problem is equivalent to showing that $$ e \leq \frac{68}{25}.$$
Method 1: Taylor expansion The above series converges rapidly, suggesting we can produce tight bounds by keeping the first several terms and then bounding the tail by a small quantity.
Expanding the series, we have
$$e = 1 + 1 + \frac{1}{2} + \frac{1}{6} + \frac{1}{24} + \frac{1}{120} + \frac{1}{720} + R ,$$
for a remainder series $R$. But the $\ell$th term of $R$ is a product of $\frac{1}{720}$ with a product of $\ell$ integers $> 2$, so $$R < \frac{1}{720} \sum_{\ell = 1}^{\infty} 2^{-\ell} = \frac{1}{720}.$$ Combining this with the previous inequality gives $e < \frac{979}{630}$, and cross-multipliying gives $\frac{979}{630} < \frac{68}{25}$. Of course, a similar method could be applied to other series with positive terms converging to $e$.
Method 2: Continued fraction expansion From the series characterization of $e$, its continued fraction expansion is $[2; 1, 2, 1, 1, 4, 1, 1, 6, \ldots, 1, 1, 2n, \ldots]$. The successive convergents of a continued fraction expansions (1) approximate the value more and more closely and (2) alternate being underestimates and overestimates. Computing from the expansion gives that the convergents of this series are: $2, 3, \frac{8}{3}, \frac{11}{4}, \frac{19}{7}, \frac{87}{32}, \ldots .$ Since $\frac{19}{7} < \frac{87}{32}$, the latter is an overestimate for $e$, and again cross-multiplying gives $e < \frac{87}{32} < \frac{68}{25}$.
A: Approach 1
As shown in this answer,
$\left(1+\frac1n\right)^n$ is increasing and $\left(1+\frac1n\right)^{n+1}$ is decreasing.
Note that $\left(1+\frac1{791}\right)^{792}=2.71999990408\lt2.72=\frac{68}{25}$
For $n\le791$, $\left(1+\frac1n\right)^n\le\left(1+\frac1{791}\right)^{791}\lt\left(1+\frac1{791}\right)^{792}\lt\frac{68}{25}$
For $n\ge791$, $\left(1+\frac1n\right)^n\lt\left(1+\frac1n\right)^{n+1}\le\left(1+\frac1{791}\right)^{792}\lt\frac{68}{25}$

Approach 2
Since $\left(1+\frac1n\right)^n$ is increasing and, as shown in this answer,
$$
\lim\limits_{n\to\infty}\left(1+\frac1n\right)^n=\sum_{n=0}^\infty\frac1{n!}
$$
it follows from the inequality,
$$
\begin{align}
\sum_{n=0}^\infty\frac1{n!}
&=\frac1{0!}+\frac1{1!}+\frac1{2!}+\frac1{3!}+\frac1{4!}\sum_{k=4}^\infty\frac{4!}{k!}\\
&\le\frac1{0!}+\frac1{1!}+\frac1{2!}+\frac1{3!}+\frac1{4!}\sum_{k=4}^\infty\frac1{5^{k-4}}\\
&=\frac1{0!}+\frac1{1!}+\frac1{2!}+\frac1{3!}+\frac1{4!}\frac54\\[6pt]
&=\frac{87}{32}\\[6pt]
&\lt\frac{68}{25}
\end{align}
$$
that $\left(1+\frac1n\right)^n\lt\frac{68}{25}$.
A: We have
$$\left(1+\dfrac1n\right)^n = \sum_{k=0}^n \dbinom{n}k \dfrac1{n^k} \leq \sum_{k=0}^n \dfrac1{k!} = 1 + \dfrac1{1!} + \dfrac1{2!} + \dfrac1{3!} + \dfrac1{4!} + \sum_{k=5}^n \dfrac1{k!}$$
For $k \geq 5$, we have
$$k! \geq 5! \times 5^{k-5} = 120 \times 5^{k-5}$$
Hence,
$$\sum_{k=5}^n \dfrac1{k!} \leq \dfrac1{120} \sum_{k=5}^{\infty} \dfrac1{5^{k-5}} = \dfrac1{120} \times \dfrac54 = \dfrac1{96}$$
Hence, we see that
$$\left(1+\dfrac1n\right)^n \leq 1 + \dfrac1{1!} + \dfrac1{2!} + \dfrac1{3!} + \dfrac1{4!} + \dfrac1{96} = 2+\dfrac{48+16+4+1}{96} = 2 + \dfrac{23}{32} \leq 2+\dfrac{18}{25}$$
A: $$
\left(1+\frac1n\right)^n < \left(1+\frac1n\right)^{n+1}
$$
and the right hand side is decreasing.
From this you get $\sup A < \left(1+\frac1{1000}\right)^{1000+1}<\frac{68}{25}$.
Edit for the non-believers: the last inequality is a true fact which can be verified with exact arithmetic: $25\times1001^{1001}<68\times1000^{1001}$.
I still don't get how this correct answer can get downvoted (but still get also copied by later answers...).
