Compute the following:
X
n≥1
n
3n−1
Remember the formula from class:
∞
X
nxn−1 =
n=1
1
(1 − x)2
All we need to do is plug in 31 into the formula to get 94 . If you forgot the formula
1
for (1−x)
2 , you can get it by differentiating the geometric series formula
∞
X
xj =
j=0
1
.
1−x
X n(n − 1)
3n−2
n≥2
This series is the same as
∞
X
n(n − 1)xn−2
n=2
evaluated at x = 13 . This series is the term-by-term derivative of
∞
X
nxn−1
n=1
So we get that
∞
X
n(n − 1)xn−2 =
n=2
Plugging in x =
1
3
gives
2
.
(1 − x)3
27
4 .
X n2
3n−2
n≥1
We can write this as
X n(n − 1)
n
+ n−2
3n−2
3
n≥1
which can, in turn, be written
X n(n − 1) n − 2
2
+ n−2 + n−2 .
3n−2
3
3
n≥1
1
We already computed the first sum to be
27
4 .
The second sum is
∞
1 X n−2
,
3 n=2 3n−3
∞
1X n
, which comes out to 34 . Finally, the
3 n=0 3n−1
∞
X
last sum can be re-indexed to 2
3n , which evaluates to 3. So the sum is
which can be re-indexed to
27
4
+
3
4
+3=
21
2 .
n=0
2