How to summarize the sigma signs

Question

I really don't know how to summarize the sigma signs in the best way, I know how to calculate with them, but to summarize them to just one sigma sign is quite new for me and I don't quite understand it, I don't really get the right answer. Thank you for helping!

$$\sum \limits _{n=1} ^k \frac {\cos (n+2)} {(n+1)^2 - 1} + \sum \limits _{l=k} ^{k+7} \left( \frac {\cos (l+3)} {(l+2)^2 - 1} -1 \right) + 8$$

Cathy

Answer 1

$$\sum_{n=1}^k{\cos(n+2)\over(n+1)^2-1}+\sum_{l=k}^{k+7}\left({{\cos(l+3)}\over {(l+2)^2-1}}-1\right)+8$$

$$\sum_{n=1}^k{\cos(n+2)\over(n+1)^2-1}+\sum_{l=k}^{k+7}\left({{\cos(l+3)}\over {(l+2)^2-1}}\right)+\sum_{l=k}^{k+7}(-1)+8$$

$$\sum_{n=1}^k{\cos(n+2)\over(n+1)^2-1}+\sum_{l=k}^{k+7}\left({{\cos(l+3)}\over {(l+2)^2-1}}\right)$$

Replace $l$ everywhere with $n-1$:

$$\sum_{n=1}^k{\cos(n+2)\over(n+1)^2-1}+\sum_{n=k+1}^{k+8}\left({{\cos(n+2)}\over{(n+2)^2-1}}\right)$$

$$\sum_{n=1}^{k+8}{\cos(n+2)\over(n+1)^2-1}$$

Answer 2

Some things to consider:

• The notation just means summing. Don't think of it as anything exotic. It is a short-hand for writing out a sum term-by-term:$$\sum_{n=1}^k \frac {\cos (n+2)} {(n+1)^2 - 1} = \frac {\cos (1+2)} {(1+1)^2 - 1} + \frac {\cos (2+2)} {(2+1)^2 - 1} + \ldots + \frac {\cos (k+2)} {(k+1)^2 - 1}$$
• Note in the above that $n$ does not occur anywhere on the right-hand-side of the equation. $n$ is a "dummy variable". It is just there to explain the notation. The value does not actually depend in any way on the value of $n$. (In fact, if $n$ has a value elsewhere in your calculation, then you really need to use a different variable than $n$ for the index.) This frees us up to use different variables for the index. As long as the function of $n$ is being summed over the same values, we can make pretty much any changes to the index we like. Note that all three sigma summations below expand out to the exact same sum of terms in a function $f$: $$\sum_{l=k}^{k+7} f(l) = \sum_{m=k}^{k+7} f(m) = \sum_{n=k+1}^{k+8} f(n-1)\\=f(k) + f(k+1)+ f(k+2)+ f(k+3)+ f(k+4)+ f(k+5)+ f(k+6)+ f(k+7)$$
• Summation is associative and commutitive, so you can rearrange a summation in any order. (I add the term "finite" because there is a concept of infinite summation that uses sigma notation in calculus. The order of summation can sometimes make a difference for it.) In particular you can break a summation up over an inner sum:$$\sum_{l=k}^{k+7} (f(l) + g(l)) = \sum_{l=k}^{k+7} f(l) + \sum_{l=k}^{k+7}g(l)$$I.e.,$$(f(k) + g(k)) + \ldots + (f(k+7) + g(k+7)) \\= (f(k) + \ldots f(k+7)) + (g(k) + \ldots + g(k+7))$$
• Simply for completeness (you don't need it here), let me also mention the distributive property: $$\sum_{l=k}^{k+7} af(l) = a\sum_{l=k}^{k+7} f(l)$$where $a$ is a constant (any expression that does not depend on the index variable $l$). This is just an application of the familiar distributive law $ab+ac = a(b + c)$.
• You can break sums up over ranges: $$\sum_{l=k}^{M} f(l) = \sum_{l=k}^{N} f(l) + \sum_{l=N+1}^{M} f(l)\\ f(k) + \ldots + f(N) + f(N+1) + \ldots + f(M) \\= (f(k) + \ldots + f(N)) + (f(N+1) + \ldots + f(M))$$where $k \le N < M$.

Except for the distributive property, you will need to use each of the concepts I've mentioned.

And I see Gerry Myerson has gone through the calculation for you while I was writing this up. I'm still posting it though, because these are concepts you should familiarize yourself with.

Answer 3

Your problem reminds me of the arabic language rooting of the word algebra, related to the medicine term mening "bone-setting" (still alive in Castillan as algebrista). You have a formula broken in three terms, and you can set them again into one.

Like the physician, before you operate, you have to look. First, the minus eight term does not seem to belong to the sums. But wait, in the second term, you get a $-1$ term, from $k$ to $k+7$, hence $(k+7) - k+1 =8$ times. Cancelled!

Now check if the bone can be set. The last term of the first expression is: $$ \frac{\cos(k+2)}{(k+1)^2-1}\,.$$ The first term of the "eight-cancelled" term is: $$ \frac{\cos(k+3)}{(k+2)^2-1}\,.$$ The two last terms are consecutive. They could be incorporated into the same series as the first term.

Its last item would be, rewritten into the first sum fashion: $$ \frac{\cos((k+7)+3)}{((k+7)+2)^2-1} = \frac{\cos((k+8)+2)}{((k+8)+1)^2-1} = \frac{\cos(n+2)}{(n+1)^2-1}\,,$$ with $n=k+8$, now that looks a lot like the first sum. Hence you get: $$ \sum_{n=1}^{k+8}\frac{\cos(n+2)}{(n+1)^2-1}\,.$$

You get even go a little further, noticing that the $n+1$ and $n+2$ can be incorporated into the sum: $$ \sum_{n=2}^{k+9}\frac{\cos(n+1)}{n^2-1}\,.$$ Now your mathematical bone is fully fixed.

In addition to the summary, you have just saved another $+$ sign and two parentheses. The curious will look at the history of the plus and the parenthesis signs at the History of mathematical notation. The history of the dagger, which looks like a $+$, is detailed in Cross footnote marker for people.