Formulas n(n + 1) k = (1)

```Formulas
n
X
n(n + 1)
2
k=
k=1
(1)
Proof. Add the sum to itself (with the terms in reverse order): 2 nk=1 k
= [1 + 2 + . . . + (n − 1) + n] + [n + (n − 1) + . . . + 2 + 1]
= [1 + n] + [2 + (n − 1)] + . . . + [(n − 1) + 2] + [n + 1] = n(n + 1).
P
n
X
1 − bn+1
, b 6= 1
b =
1−b
k=0
k
Pn
Proof. Expand the product (1 − b)
k=0
∞
X
bk =
k=0
Proof. Take the limit of
Pn
k=0
bk =
1−bn+1
1−b
n
X
Pn
bk =
k=0
(2)
Pn+1 k
b
bk −
k=1
= b0 − bn+1 .
1
, |b| &lt; 1
1−b
(3)
as n → ∞.
ln k = Θ(n ln n)
(4)
k=0
Proof.
First note that the sum isR an approximation sum for an integral: a lower sum for
Rn
n
1 ln xdx and an upper sum for 2 ln xdx. It follows that
Z n
ln xdx &lt;
2
n
X
ln k &lt;
Z n
ln xdx
1
1
Thus n1 ln k = Θ( 1n ln xdx). Evaluation of the integral by parts yields
(n − 1) = Θ(n ln n).
P
R
Rn
1
ln xdx = n ln n −
n
X
1
= Θ(ln n)
k
k=1
(5)
Proof. As in the argument above,
the sum is a lower approximation
sum for 1n (1/x)dx
Rn
Rn
and an upper approximation for 2 (1/x)dx,
hence the sum is Θ( 1 dx/x). But this integral
R
is the definition of natural logarithm: 1n dx/x = ln n
R
ln n! = Θ(n ln n)
(6)
√
Proof. Applying Sterling’s approximation, we have ln n! = log( 2πn) + n ln(n/e) + ln(1 +
Θ(1/n)) = Θ(n ln n).
√
n n
1
e
n
Proof. This is Sterling’s approximation - proof beyond the scope of this page.
n! =
2πn
1
1+Θ
(7)
```