SOME REMARKS ON THE SUM OF AN OLD SERIES

Rasul A. Khan

In this note we use some combinatorial identities to derive a formula for the sum of the series

S(p) = ∑_{n=0}^{∞}(m + n + 1)^{p}\binom{n + m}{m}x^{n}, |x| < 1,

in the form P (x)/(1 − x)^{m+p+1}, where P (x) is a polynomial of degree p − 1 with known coefficients a_{j}, 0 ≤ j ≤ p − 1. When specialized for m = 0, the resulting sum gives a formula for

∑_{n=1}^{∞} n^{p}x^{n} (|x| < 1).

The general formula also provides an alternative method for determining the moments of a negative binomial distribution. Conversely, the negative binomial distribution can be used to find a recursive formula for the sum of the above series S(p).

1. A Combinatorial Identity. In what follows we write

x_{(r)} = x(x − 1)(x − 2) ⋯ (x − r + 1)

for any real x and positive integer r, and in particular x_{(r)} = x!/(x − r)! if x is also a positive integer, and x_{(0)} = 1, etc. Also, for a function ψ(t), ψ^{(k)}(a) denotes the kth derivative evaluated at a. The identities in Lemma 1 can be found in disguised forms in Feller [2]. These identities in turn imply the main identity in Lemma 2. A generating function method is used for their proofs for the sake of completeness.

\underline{Lemma 1}. For any real number m and positive integers n and r we have

(a) ∑_{j=0}^{n}(−1)^{j}(m − j)^{r}\binom{n}{j} = \{ \begin{matrix}0, & for all r < n, \\ n!, & for r = n. \end{matrix}

(b) ∑_{j=0}^{n}(−1)^{j}\binom{n}{j}(n + m − j)_{(r)} = \{ \begin{matrix}0, & for all r < n, \\ n!, & for r = n. \end{matrix}

\underline{Proof}. Let

g(t) = e^{mt}(1 − e^{−t})^{n} = ∑_{j=0}^{n}(−1)^{j}\binom{n}{j} exp((m − j)t).

Since g^{(r)}(0) = 0 for all r < n, and g^{(n)}(0) = n!, (a) follows. Now let

g_{1}(t) = t^{m}(t − 1)^{n} = ∑_{j=0}^{n}(−1)^{j}\binom{n}{j}t^{m+n−j}.

Since g_{1}^{(r)}(1) = 0 for all r < n, and g^{(n)}(1) = n!, (b) follows.

To state the second lemma we define the following sequence a_{n} for positive integers m and p. This sequence is needed to find the sum of the series S(p). The sequence a_{n} is defined as:

a_{n} = ∑_{i=0}^{n}(−1)^{i}(m + n + 1 − i)^{p}\binom{m + p + 1}{i}\binom{n + m − i}{m}.  (1)

\underline{Lemma 2}.

∑_{j=0}^{n}(−1)^{j}(m + n + 1 − j)^{p}\binom{n}{j}(m + n − j)_{(r)} = 0, r = n − p − 1.

Moreover, a_{n} = 0 for all n ≥ p, where a_{n} is defined by (1).

\underline{Proof}. Let g(t) = t^{m} exp((m + 1)(t − 1))(te^{t−1} − 1)^{n}. Clearly,

g(t) = ∑_{j=0}^{n}(−1)^{j}\binom{n}{j}t^{m+n−j} exp((m + n + 1 − j)(t − 1)).  (2)

Let f_{1}(t) = t^{m+n−j} and f_{2}(t) = exp((m + n + 1 − j)(t − 1)). Obviously,

f_{1}^{(r)}(1) = (m + n − j)_{(r)}, r = n − p − 1, and f_{2}^{(p)}(1) = (m + n + 1 − j)^{p}.

It is easy to verify that g^{(k)}(1) = 0 for all k < n. By repeated differentiation of (2) and repeated use of Lemma 1 we obtain

∑_{j=0}^{n}(−1)^{j}\binom{n}{j}f_{1}^{(α)}(1)f_{2}^{(β)}(1) = 0, α + β = k < n.  (3)

In particular, (3) holds if α = n − p − 1 and β = p, and the identity is proved. To prove the noted property of a_{n} we consider two separate cases. If n = p, we have

(m + p + 1 − i)^{p}\binom{m + p + 1}{i}\binom{p + m − i}{m} = \frac{(m + p + 1)!}{m!p!}(m + p + 1 − i)^{p−1}\binom{p}{i},

and a_{p} = 0 by Lemma 1(a). If n > p (i.e. n ≥ p + 1) we write

\binom{m + p + 1}{i}\binom{n + m − i}{m} = \frac{(m + p + 1)!}{m!n!}\binom{n}{i}(n + m − i)_{(r)}, r = n − p − 1.

Then, “showing a_{n} = 0 for all n > p” is equivalent to the main identity in Lemma 2. Hence, a_{n} = 0 for all n ≥ p.

2. An Application. The mth derivative of the geometric series

∑_{n=0}^{∞} x^{n} = (1 − x)^{−1}, |x| < 1,

gives the well–known sum [4]

∑_{n=0}^{∞} \binom{n + m}{m}x^{n} = \frac{1}{(1 − x)^{m+1}} , |x| < 1.  (4)

If 0 < x < 1, then

f(n; m, x) = \binom{n + m}{m}(1 − x)^{m+1}x^{n}, n = 0, 1, 2, …

is the probability function of a negative binomial distribution, and (4) can be interpreted as

∑_{n=0}^{∞} f(n; m, x) = 1 (cf. [2]).

Motivated by this observation and the related moments problem we seek to derive a formula for

∑_{n=0}^{∞}(m + n + 1)^{p}\binom{n + m}{m}x^{n}, |x| < 1,

where m (≥ 0) and p (≥ 0) are integers. It is claimed that

∑_{n=0}^{∞}(m + n + 1)^{p}\binom{n + m}{m}x^{n} = \frac{P (x)}{(1 − x)^{m+p+1}} , |x| < 1,  (5)

where

P (x) = ∑_{j=0}^{p−1} a_{j}x^{j}

with known coefficients a_{j} (0 ≤ j ≤ p − 1), and P (x) = 1 if p = 0. To see this, let

b_{n} = (m + n + 1)^{p}\binom{n + m}{m}, c_{n} = (−1)^{n}\binom{m + p + 1}{n},

and consider the power series

A = ∑_{n=0}^{∞} b_{n}x^{n}, and B = (1 − x)^{m+p+1} = ∑_{n=0}^{m+p+1} c_{n}x^{n}.

Then

AB = ∑_{n=0}^{∞} a_{n}x^{n}

by the Cauchy product formula, and for n = 0, 1, 2, …, the coefficients a_{n} are given by (1). Since a_{n} = 0 for all n ≥ p by Lemma 2,

AB = ∑_{n=0}^{p−1} a_{n}x^{n} = P (x), and A = P (x)/B,

and (5) follows.

\underline{Special Case}. Put m = 0 in (5), and the formula becomes

∑_{n=0}^{∞}(n + 1)^{p}x^{n} = ∑_{n=1}^{∞} n^{p}x^{n−1} = \frac{P (x)}{(1 − x)^{p+1}} ,  (6)

where

P (x) = ∑_{j=0}^{p−1} a_{j}x^{j},

and a_{n} in (1) reduces to

a_{n} = ∑_{i=0}^{n}(−1)^{i}(n + 1 − i)^{p}\binom{p + 1}{i}.

Clearly, a_{0} = 1 and a_{n} = 0 for all n ≥ p. In this special case there is one more interesting observation, namely, a_{p−1} = 1 for all p ≥ 1. This can be seen by noting that

\begin{aligned}a_{p−1} & = ∑_{i=0}^{p−1}(−1)^{i}(p − i)^{p}\binom{p + 1}{i} = ∑_{i=0}^{p+1}(−1)^{i}(p − i)^{p}\binom{p + 1}{i} − (−1)^{p+1}(−1)^{p} \\ & = ∑_{i=0}^{p+1}(−1)^{i}(p − i)^{p}\binom{p + 1}{i} + 1. \end{aligned}

The last sum is clearly zero by Lemma 1(a), and hence, a_{p−1} = 1 for all p ≥ 1. Obviously, (6) gives a formula for

∑_{n=1}^{∞} n^{p}x^{n}

as

\frac{xP (x)}{(1 − x)^{p+1}} ,

which was also noted by Clarke [1] and Stalley [4]. Another curious observation is worth mentioning. Since P (x) is a polynomial of degree p − 1 and the a_{j}’s are integers,

S_{p} = ∑_{n=1}^{∞} \frac{n^{p}}{2^{n}} = 2^{p}P (\frac{1}{2})

is always an even integer. It follows from (5) that

S_{p}(m) = ∑_{n=1}^{∞}(n + m)^{p}\binom{n + m − 1}{m}\frac{1}{2^{n}} = 2^{p+m}P (\frac{1}{2})

is also an even integer for any m ≥ 0 and p ≥ 0. Clearly, S_{p} = 2^{p} + 2^{p−1}a_{1} + 2^{p−2}a_{2} + ⋯ + 2. In this special case one can easily check that S_{1} = 2, p = 2, a_{1} = 1, S_{2} = 6, p = 3, a_{1} = 4, a_{2} = 1, S_{3} = 26, p = 4, a_{1} = 11, a_{2} = 11, a_{3} = 1, S_{4} = 150, and p = 5, a_{1} = 26, a_{2} = 66, a_{3} = 26, a_{5} = 1, S_{5} = 1082, etc.

3. A Probabilistic Approach and a Recursive Formula for S(p). Let 0 < x < 1, and

P (ν = n) = f(n; m, x) = \binom{n + m}{m}(1 − x)^{m+1}x^{n}, n = 0, 1, 2, …

be the probability function of a negative binomial distribution (cf. [2]). Let X = ν + m + 1. Then the pth moment of X can be obtained from (5), which is given by

EX^{p} = E(ν + m + 1)^{p} = ∑_{n=0}^{∞}(n + m + 1)^{p}P (ν = n) = \frac{P (x)}{(1 − x)^{p}} .  (7)

In particular, if p = 1, then a_{0} = m + 1, and EX = (m + 1)/(1 − x), and if p = 2, then a_{0} = (m + 1)^{2}, a_{1} = m + 1, and

EX^{2} = \frac{(m + 1)^{2} + (m + 1)x}{(1 − x)^{2}}.

Moreover, it is obvious that the variance σ^{2} = E(X − EX)^{2} = EX^{2} − (EX)^{2}) = (m+1)x/(1−x)^{2}. These formulas are known in the literature where it is customary to use q for x and p for 1 − x, and r = m + 1. Of course, any moment of X can be calculated from (7).

There is an alternative derivation of the sum

S(p) = ∑_{n=0}^{∞}(n + m + 1)^{p}\binom{n + m}{m}x^{n}

by calculus via the negative binomial distribution. This interesting interplay between S(p) and the negative binomial distribution is now discussed. This method produces a recursive formula for S(p). For simplicity we write r = m + 1 (m being a positive integer) in the above negative binomial distribution. Thus,

f(n; r, x) = (1 − x)^{r}\binom{r + n − 1}{r − 1}x^{n}, n = 0, 1, 2, …,

and the associated moment generating function ϕ^{∗}(θ) = E exp(θX) is given by

ϕ^{∗}(θ) = ∑_{n=0}^{∞} e^{nθ}f(n; r, x) = \frac{(1 − x)^{r}}{(1 − xe^{θ})^{r}} = \frac{1}{(1 − β(e^{θ} − 1))^{r}} , θ < − ln x,

where β = x/(1 − x). Consequently,

ϕ(θ) = E exp(θ(X + r)) = exp(rθ)/(1 − β(e^{θ} − 1))^{r},

and ϕ′(θ) = r(1 + β)ϕ(θ)/(1 − β(e^{θ} − 1)). Thus,

ϕ′(θ)(1 − β(e^{θ} − 1)) = r(1 + β)ϕ(θ).  (8)

Differentiating (8) p − 1 (p ≥ 1) times, by Leibniz’s formula we have

∑_{j=0}^{p−1} \binom{p − 1}{j}ϕ^{(j+1)}(θ)(1 − β(e^{θ} − 1))^{(p−1−j)} = r(1 + β)ϕ^{(p−1)}(θ),

which evaluated at θ = 0 gives

−β ∑_{j=0}^{p−2} \binom{p − 1}{j}ϕ^{(j+1)}(0) + ϕ^{(p)}(0) = r(1 + β)ϕ^{(p−1)}(0).

This can be written as

−β ∑_{j=1}^{p−1} \binom{p − 1}{j − 1}ϕ^{(j)}(0) + ϕ^{(p)}(0) = r(1 + β)ϕ^{(p−1)}(0).

Since

\binom{p − 1}{j − 1} = \frac{j}{p}\binom{p}{j},

we obtain

ϕ^{(p)}(0) = β ∑_{j=1}^{p−1} \frac{j}{p}\binom{p}{j}ϕ^{(j)}(0) + r(1 + β)ϕ^{(p−1)}(0)

= β ∑_{j=1}^{p−1} \frac{p − p + j}{p} \binom{p}{j}ϕ^{(j)}(0) + r(1 + β)ϕ^{(p−1)}(0)

= β ∑_{j=1}^{p−1} \binom{p}{j}ϕ^{(j)}(0) − β ∑_{j=1}^{p−1} \frac{p − j}{p} \binom{p}{j}ϕ^{(j)}(0) + r(1 + β)ϕ^{(p−1)}(0).

Since

\frac{p − j}{p} \binom{p}{j} = \binom{p − 1}{j},

we have

ϕ^{(p)}(0) = β ∑_{j=1}^{p−1} \binom{p}{j}ϕ^{(j)}(0) − β ∑_{j=1}^{p−1} \binom{p − 1}{j}ϕ^{(j)}(0) + r(1 + β)ϕ^{(p−1)}(0). (9)

This recursive formula can be used for computing moments E(X+r)^{p}. Equivalently, it can be used to find the sum

S(p) = ∑_{n=0}^{∞}(n + r)^{p}\binom{n + r − 1}{r − 1}x^{n},

and since r = m + 1,

S(p) = ∑_{n=0}^{∞}(n + m + 1)^{p}\binom{n + m}{m}x^{n}.

It is clear from the definition of the generating function ϕ(θ) that S(p) = ϕ^{(p)}(0)/(1 − x)^{r} for each p ≥ 0. Dividing equation (9) by (1 − x)^{r} we obtain

S(p) = β ∑_{j=1}^{p−1} \binom{p}{j}S(j) − β ∑_{j=1}^{p−1} \binom{p − 1}{j}S(j) + r(1 + β)S(p − 1),

which can be written as

S(p) = β(1 + ∑_{j=1}^{p−1} \binom{p}{j}S(j)) − β(1 + ∑_{j=0}^{p−1} \binom{p − 1}{j}S(j)) + r(1 + β)S(p − 1). (10)

This can be used to find S(p) in a recursive manner for any desired p. Conversely, (10) can also be used to determine the moments of the negative binomial distribution. Suppose we want to determine

S_{1}(p) = xS(p) = ∑_{n=0}^{∞}(n + m + 1)^{p}\binom{n + m}{m}x^{n+1}.

It is obvious from (10) that the same recursive equation holds for S_{1}(p) also. An interesting recursive equation arises if m = 0 (i.e., r = 1). In this case (10) becomes

S_{1}(p) = β(1 + ∑_{j=0}^{p−1} \binom{p}{j}S_{1}(j)) − β(1 + ∑_{j=0}^{p−1} \binom{p − 1}{j}S_{1}(j)) + (1 + β)S_{1}(p − 1).

This can be written as

S_{1}(p) = β(1 + ∑_{j=0}^{p−1} \binom{p}{j}S_{1}(j)) − β(1 + ∑_{j=0}^{p−2} \binom{p − 1}{j}S_{1}(j)) + S_{1}(p − 1), (11)

where

S_{1}(p) = ∑_{n=0}^{∞}(n + 1)^{p}x^{n+1} = ∑_{n=1}^{∞} n^{p}x^{n}.

Note that (11) implies

S_{1}(p) = β(1 + ∑_{j=0}^{p−1} \binom{p}{j}S_{1}(j))  (12)

by induction. The recursive formula (12) was derived in [3] by the telescoping method. If x = 1/2, then β = 1, and since S_{1}(0) = 1, it is found that S_{1}(1) = 2, S_{1}(2) = 6, S_{1}(3) = 26, S_{1}(4) = 150, S_{1}(5) = 1082, etc. which were obtained in the preceding section by the combinatorial method.

\underline{References}

1. E. H. Clarke, “A Formula for the Sum of a Certain Type of Infinite Power Series,” American Mathematical Monthly, 21 (1914), 292–296.

2. W. Feller, An Introduction to Probability Theory and its Applications, Wiley, New York, 1968.

3. R. A. Khan, “Another Elementary Proof of the Convergence-Divergence of the p-Series,” Missouri Journal of Mathematical Sciences, 16 (2004), 104–107.

4. K. Knopp, Theory and Applications of Infinite Series, Blackie & Son Ltd., London, 1928.

5. R. Stalley, “A Generalization of the Geometric Series,” American Mathematical Monthly, 56 (1949), 325–327.

Mathematics Subject Classification (2000): 40A25, 65B10; 40A99

Rasul A. Khan Department of Mathematics Cleveland State University Cleveland, OH 44115 email: khan@math.csuohio.edu