Binomial theorem

$$ (x+y)^n = \sum_{k=0}^n {n \choose k} x^{n - k} y^k $$

Exponential function

$$ e^x = \lim_{n \to \infty} \left( 1+ \frac{x}{n} \right)^n $$

Cauchy–Schwarz inequality

$$ \left( \sum{k=1}^n ak bk \right)^2 \leq \left( \sum{k=1}^n ak^2 \right) \left( \sum{k=1}^n b_k^2 \right) $$

Bayes' theorem

$$ P(A \mid B) = \frac{P(B \mid A) \, P(A)}{P(B)} $$

Euler's summation formula

Theorem. Euler's summation formula. If $f$ has a continuous derivative $f'$ on the interval $[y, x]$, where $0 < y < x$, then

\begin{align} \label{theorem} \sum{y < n \le x} f(n) = & \inty^x f(t) dt + \int_y^x (t - [t]) f'(t) dt

otag \ & + f(x)([x] - x) - f(y)([y] - y). \end{align}

Proof. Let $m = [y]$, $k = [x]$. For integers $n$ and $n - 1$ in $[y, x]$ we have \begin{align} \int{n-1}^n [t] f'(t) dt & = \int{n-1}^n f'(t) dt \ & = (n - 1) \bigl( f(n) - f(n - 1) \bigr) \ & = \bigl( n f(n) - (n - 1) f(n - 1) \bigr) - f(n). \end{align} Summing from $n = m + 1$ to $n = k$ we find \begin{align} \int{m}^k [t] f'(t) dt & = \sum{n = m + 1}^k \bigl( n f(n) - (n - 1) f(n - 1) \bigr) - \sum{y < n \le x} f(n) \ & = k f(k) - m f(m) - \sum{y < n \le x} f(n). \end{align} Hence, \begin{align} \label{summation} \sum{y < n \le x} f(n) & = - \int{m}^k [t] f'(t) dt + k f(k) - m f(m)

otag \ & = - \int{y}^x [t] f'(t) dt + k f(x) - m f(y). \end{align} Integration by parts gives us \begin{equation*} \inty^x f(t) dt = x f(x) - y f(y) - \int_y^x t f'(t) dt. \end{equation*} When this is combined with \eqref{summation} we obtain \eqref{theorem}.