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 $$\frac{68}{25}\lt(1+\frac{1}{n})^n\Rightarrow\frac{(68)^{1/n}}{(25)^{1/n}}-\frac{1}{n}&lt;1\Rightarrow n(68)^{1/n}-(25)^{1/n}\lt n(25)^{1/n}\Rightarrow n(68)^{1/n}\lt(25)^{1/n}(n+1)$$ At this part i get stuck, because, we see that $n&lt;n+1$ and instinctively i'd say that

$(68)^{1/n}&lt;(25)^{1/n}$, apparently, i´ve found a contradiction: $68&lt;25$. But i've found that in general if

we have $a&lt;b$ and $ac&lt;bd$ it´s not always true that $c&lt;d$, for instance, $2&lt;3$ and $2\sqrt2&lt;3(1)$

don't imply that 
$\sqrt2&lt;1$. What can i do?, Please, help me. THANKS
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 $&gt; 2$, so $$R &lt; \frac{1}{720} \sum_{\ell = 1}^{\infty} 2^{-\ell} = \frac{1}{720}.$$ Combining this with the previous inequality gives $e &lt; \frac{979}{630}$, and cross-multipliying gives $\frac{979}{630} &lt; \frac{68}{25}$. 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 $\frac{19}{7} &lt; \frac{87}{32}$, the latter is an overestimate for $e$, and again cross-multiplying gives $e &lt; \frac{87}{32} &lt; \frac{68}{25}$.
$$
\left(1+\frac1n\right)^n &lt; \left(1+\frac1n\right)^{n+1}
$$
and the right hand side is decreasing.
From this you get $\sup A &lt; \left(1+\frac1{1000}\right)^{1000+1}&lt;\frac{68}{25}$.

Edit for the non-believers: the last inequality is a true fact which can be verified with exact arithmetic: $25\times1001^{1001}&lt;68\times1000^{1001}$.

I still don't get how this correct answer can get downvoted (but still get also copied by later answers...).