Telescoping series of form $\sum (an+1) ..(an+k)$

Question

If we are given a series say

$$\sum_{n=1}^N (n+1)(n+2) ... (n+k)$$ we can find a telescoping series by noting that

$(n+1)(n+2)..(n+k)(n+k+1) - n(n+1)..(n+k) = (n+k+1-n) (n+1)...(n+k)$

and hence able to write

$$\sum_{n=1}^N (n+1)(n+2) ... (n+k) = \frac{1}{k+1}\sum_{n=1}^N (n+1)...(n+k+1) - n(n+1)...(n+k)$$

However, for series like $$\sum_{n=1}^N (2n+1)(2n+2) ... (2n+k)$$ or for a general multiple of $n$, such as $$\sum_{n=1}^N (an+1)(an+2) ... (an+k)$$ the same trick to add a term before and after would not work.

Are there any ways to express series like this as a telescoping sum, or failing that, any useful (relatively) closed form identities? If so, a pointer to a reference would be much appreciated!

Answer 1

The original identity comes from finding an antidifference for the function $n^{\underline{k}}$. Unfortunately, no such antidifference exists for $(2n)^{\underline{k}}$, due to lack of a "chain rule" for antidifferences. However here's an approach that you might like:

$$a_k=\sum_{n=1}^N(2n+1)(2n+2)\cdots(2n+k)$$ $$b_k=\sum_{n=1}^N (2n)(2n+1)\cdots(2n+k-1)$$

Multiplying the summands of $a_k$ by $(2n+k+1-2n)=k+1$, we get $a_k=\frac{1}{k+1}(a_{k+1}-b_{k+1})$. Multiplying the summands of $b_k$ by $(2n+k-2n+1)=k+1$, we get $b_k=\frac{1}{k+1}(b_{k+1}-\sum_{n=0}^{N-1}(2n+1)(2n+2)\cdots(2n+k+1))=\frac{1}{k+1}(b_{k+1}-a_{k+1}+s_k)$, where $s_k=(2N+1)(2N+2)\cdots(2N+k+1)-(k+1)!$.

Adding, we get $$a_k+b_k=\frac{s_k}{k+1}$$

Replacing $k$ by $k+1$ and rearranging, we get $$-b_{k+1}=a_{k+1}-\frac{s_{k+1}}{k+2}$$

We plug into our above formula for $a_k$ to get $a_k=\frac{1}{k+1}(2a_{k+1}-\frac{s_{k+1}}{k+2})$. Now we may solve for $a_{k+1}$ to get $$a_{k+1}=\frac{k+1}{2}a_k+\frac{s_{k+1}}{2k+4}$$ which admittedly is no closed form, but it is a first-order recurrence.

Answer 2

Let $P(k,a,j) =\prod_{i=1}^k (ai+j) =a^k \prod_{i=1}^k (i+j/a) $.

Instead of comparing the next term with the current term, I will compare the term $a$ further on (which is the next term when $a=1$).

$\begin{align} P(k, a, j+a) - P(k, a, j) &=a^k \prod_{i=1}^k (i+(j+a)/a) -a^k \prod_{i=1}^k (i+j/a)\\ &=a^k\big( \prod_{i=1}^k (i+(j+a)/a) -\prod_{i=1}^k (i+j/a)\big)\\ &=a^k\big( \prod_{i=1}^k (i+1+j/a) -\prod_{i=1}^k (i+j/a)\big)\\ &=a^k\big( \prod_{i=2}^{k+1} (i+j/a) -\prod_{i=1}^k (i+j/a)\big)\\ &=a^k \big(\prod_{i=2}^{k} (i+j/a)\big) \big ((k+1+j/a) - (1+j/a)\big)\\ &=a^k k\prod_{i=2}^{k} (i+j/a)\\ &=a k\prod_{i=2}^{k} (ai+j)\\ &=a k\prod_{i=1}^{k-1} (a(i+1)+j)\\ &=a k\prod_{i=1}^{k-1} (ai+a+j)\\ &=a kP(k-1, a, j+a)\\ \end{align} $

or $P(k+1, a, j+a) - P(k+1, a, j) =a(k+1) P(k, a, j+a) $ or $\dfrac{P(k+1, a, j+a) - P(k+1, a, j)}{a(k+1)} = P(k, a, j+a) $ or $\dfrac{P(k+1, a, j) - P(k+1, a, j-a)}{a(k+1)} = P(k, a, j) $

For $a=1$ this becomes $P(k+1, 1, j+1) - P(k+1, 1, j) =(k+1)P(k-1, 1, j+1) $ or $P(k+1, 1, j) - P(k+1, 1, j-1) =(k+1)P(k-1, 1, j) $.

So (note the limits of summation)

$\begin{align} \sum_{j=1}^{Na} P(k, a, j) &=\frac1{a(k+1)}\sum_{j=1}^{Na} \big(P(k+1, a, j) - P(k+1, a, j-a)\big)\\ &=\frac1{a(k+1)}\big(\sum_{j=1}^{Na} P(k+1, a, j) -\sum_{j=1}^{Na} P(k+1, a, j-a)\big)\\ &=\frac1{a(k+1)}\big(\sum_{j=Na-a+1}^{Na} P(k+1, a, j) -\sum_{j=1-a}^{0} P(k+1, a, j)\big)\\ \end{align} $

We have $a$ terms which involve $N$ and $a$ terms which do not (and so are constant).

If the upper limit is not a multiple of $a$, then there are from $1$ to $a-1$ terms of the form $P(k, a, Na+i)$.

This is a reasonable generalization (IMHO) of the telescoping sum for $a=1$.