Why $e>n^{3/2}$ prevents Wiener's attack Here's the Wiener's attack as I understand it.

Suppose $n=pq$ with $q &lt; p &lt; 2q$, and $d &lt; n^{1/4}/3$ where $ed=k\phi(n)+1$ and $e &lt; \phi(n)$.

$$\begin{align*}
n-\phi(n) &amp;=n-(p-1)(q-1)\\
&amp;= n-(n-p-q+1) \\
&amp;= p+q-1 \\
&amp;&lt; 2q+q-1\\
&amp;=3q-1 \\
&amp;&lt; 3\sqrt{n}
\end{align*}$$

and $e&lt;\phi(n)\implies ke &lt; k\phi(n) =ed -1 &lt; ed \implies k&lt;d&lt;n^{1/4}/3$.

Therefore, 
$$\begin{align*}
\left|\frac{e}{n}-\frac{k}{d}\right| &amp;= \left|\frac{ed-kn}{nd}\right|\\
&amp;=\left|\frac{de-k\phi(n)-(kn+k\phi(n))}{nd}\right|\\
&amp;= \left|\frac{1-k(n-\phi(n))}{nd}\right|\\
&amp;\le \frac{3k\sqrt{n}}{nd} \\
&amp;\le \frac{3(n^{1/4}/3)\sqrt{n}}{nd} = \frac{1}{dn^{1/4}}\\
&amp;&lt; \frac{1}{2d^2}.
\end{align*}$$

The last inequality (by some theorem about continued fraction) and $gcd(d,k)=1$ imply that $\frac{k}{d}$ is a convergent in simple continued fraction of $\frac{e}{n}$.

Why does $e&gt;n^{3/2}$ imply that $k/d$ is not a convergent in simple continued fraction of $e/n$? Or why does such an $e$ prevent Wiener's attack?

It suffices to show that $|e/n-k/d| &gt; 1/d^2$.

Thank you!
In Wiener's attack $k$ is a small number ($k&lt;d$). If we set $e^{'}=e+t\phi(n)$ for some large $t$ then $e^{'}$ can be used in place of $e$ for message encryption.Also the k became a big number and we can't using of Wiener's attack.For more detail you can see "Twenty Years of Attacks on the RSA Cryptosystem".