I'm working on a problem and I'm not sure how to find the product of these two sums:

$\left(\sum_{k=0}^{\infty}\text{something}\right)\left(\sum_{k=n}^{n}\text{something else}\right)$

The "something" and "something else" are something more specific for my problem, but I really just need help wrapping my head around finding the product of two sums where one is finite and one is infinite. Would something to do with convolutions help? I'm just not sure. Thanks in advance.

Edit: The "somethings" have been added in the comments.

  • $\begingroup$ As you've written these, both are infinite. Did you mean something different? Also, you'll likely find yourself confused if you use $k$ as the index of summation for two different summations. $\endgroup$ – davidlowryduda Jan 27 '16 at 7:41
  • $\begingroup$ How well do the factor sums converge? $\endgroup$ – Hagen von Eitzen Jan 27 '16 at 7:43
  • $\begingroup$ Yes, sorry. the second sum should just be to n. just the first is infinite. $\endgroup$ – User Jan 27 '16 at 7:47
  • $\begingroup$ I'll give you the actual sums, maybe it'll make more sense: $\left( \sum_{k=0}^{\infty} {{n+k-1}\choose{k}}x^k\right)\left(\sum_{k=0}^{n} {{n}\choose{k}}(-1)^kx^k \right)$ @mixedmath $\endgroup$ – User Jan 27 '16 at 7:47
  • $\begingroup$ @hagenvoneitzen I added the actual "somethings" in the comments. Perhaps that will make a difference? $\endgroup$ – User Jan 27 '16 at 7:51

You have a product of shape $$ \sum_{k = 0}^\infty a_k \sum_{\ell = 0}^n b_\ell. \tag{1}$$ You will notice that I've done a few things differently than you. Firstly, I've given the two (somethings) names. Secondly, and more importantly, I've indicated that the first sum is over $k$ and the second sum is over $\ell$. It is important to not reuse indices (like $k$) for different sums in the same expression in order to retain well-definedness.

The product is not exciting. In short, the product is written in the exact same way as one might think of $$ (a_1 + a_2 + a_3)(b_1 + b_2 + b_3) = a_1b_1 + a_1b_2 + a_1b_3 + \cdots + a_3b_1 + a_3b_2 + a_3b_3,$$ in that the product is each possible choice of first multiplicand and second multiplicand, added together.

One way to write $(1)$ is to write $$ \sum_{k = 0}^\infty \sum_{\ell = 0}^n a_k b_\ell.$$ This is not exciting, but it is the product; and it is exactly the sum over choices of first and second multiplicand.

I suspect you mean something a little bit different. So let's now think about sums of shape $$ \sum_{k = 0}^\infty a_k x^k \sum_{\ell = 0}^n b_\ell x^\ell.$$ It is a very reasonable question to wonder what the coefficient of $x^m$ is in this product. So let's first rewrite the product, $$ \sum_{k = 0}^\infty \sum_{\ell = 0}^n a_k b_\ell x^{k + \ell}.\tag{2}$$ The coefficient of $x^m$ comes from those terms with an $x^m$ appearing, which happen when $k + \ell = m$.

When can $k + \ell = m$? you are summing over all natural numbers $k$ and those natural numbers $\ell$ up to $n$. So for $m \leq n$, we can think of pairing $\ell$ with $k = m - \ell$ to see that the coefficient of $x^m$ in $(2)$ is $$ \sum_{k + \ell = m} a_k b_\ell = \sum_{\ell = 0}^m a_{m - \ell} b_{\ell}.$$ For $m > n$, we only have $n$ terms. We can still pair $\ell$ with $k = m - \ell$, but only for $\ell$ up to $n$. So for $m > n$, the coefficient of $x^m$ in $(2)$ is $$ \sum_{k + \ell = m}' a_k b_\ell = \sum_{\ell = 0}^n a_{m - \ell}b_\ell.$$

Aside: You will notice that I've used a primed summation symbol --- this is to indicate that there is something strange about the sum. Here, I mean that there is a restriction on $\ell$ which I did not make explicit in the first sum. Sometimes, people are more explicit, and they might write it as $$ \sum_{\substack{k + \ell = m \\ 0 \leq \ell \leq n}} a_k b_\ell.$$ But if you have never seen conditions below indices of summation (like I expect that the OP has not seen these before), this can look intimidating.

So you could regroup all of these together and write the sum as $$ \sum_{m = 0}^n \sum_{\ell = 0}^m a_{m - \ell} b_\ell x^m + \sum_{m = n+1}^\infty \sum_{\ell = 0}^n a_{m - \ell} b_\ell x^m.$$ Or, more compactly but less legibly, you could write it as $$ \sum_{m = 0}^\infty \sum_{\ell = 0}^{\min\{m,n\}} a_{m - \ell} b_\ell x^m,$$ where $\min\{m, n\}$ denotes the minimum of $m$ and $n$.

  • 1
    $\begingroup$ Thank you so much! This answer was very thorough and it helped me get to the next step of me solving the rest of the problem (and I think I got the rest from here!). I appreciate the quick help. $\endgroup$ – User Jan 27 '16 at 8:24
  • $\begingroup$ This part in particular was very clever: $\sum_{k + \ell = m} a_k b_\ell = \sum_{\ell = 0}^m a_{m - \ell} b_{\ell}.$ $\endgroup$ – User Jan 27 '16 at 8:41

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