update notations and fix typos

pages/Math4111/Math4111_L19.md

@@ -15,14 +15,14 @@
     &\geq\binom{n}{4}
     \end{aligned}
     $$
-    EOP
+    QED
 2. Using part 1, show that $\lim_{n\to\infty}\frac{n^3}{2^n}=0$.  
    Proof:
    $$
    \frac{n^3}{2^n}\leq\frac{n^3}{\binom{n}{4}}
    $$
    The value of $\frac{n^3}{\binom{n}{4}}$ is decreasing when $n\geq 4$.
-   EOP
+   QED
 
 ## New materials
 
@@ -53,7 +53,7 @@ $$
 |s_m-s_n|=\left|\sum_{k=n}^{m}a_k\right|<\epsilon.
 $$
 
-EOP
+QED
 
 Special case of this theorem.
 
@@ -94,7 +94,7 @@ $$
 \left|\sum_{k=n}^{m}c_k\right|\leq \sum_{k=n}^{m}c_k<\epsilon.
 $$
 
-EOP
+QED
 
 #### Theorem 3.26 (Geometric series)
 
@@ -115,7 +115,7 @@ So $s_n=\frac{1-x^{n+1}}{1-x}$.
 
 Since $|x|<1$, $x^{n+1}$ converges to 0. So $\lim_{n\to\infty}s_n=\frac{1}{1-x}$.
 
-EOP
+QED
 
 #### Lemma 3.28 
 
@@ -149,7 +149,7 @@ $$
 
 > Fun fact: $\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}$.
 
-EOP
+QED
 
 #### Theorem 3.27 (Cauchy condensation test)
 
@@ -191,4 +191,4 @@ So $(s_n)_{n=1}^{\infty}$ is a bounded above.
 
 By **Theorem 3.14**, $(s_n)_{n=1}^{\infty}$ converges if and only if $(t_k)_{k=0}^{\infty}$ converges.
 
-EOP
+QED