Does this sigma function correspond to this series? I'm kind of self-taught in sigmas, and I'm not very proficient in calculus nor precalculus. Therefore, I'd like to see if the series I've thought of does correspond to this sigma function
$$\sum_{k=1}^n k\times\frac k2$$
Corresponds to this series:
$$1\times\frac 12+2\times \frac 22+3\times \frac 32 + 4\times\frac 42 ... n\times\frac n2$$
I'm wondering because I want to do a little recreational study of this series, and see if what kind of relations n has to the sums. To do that, it would be nice to compress that series to a sigma function, and I'm wondering if I did it correctly here.
Adding this as an answer to avoid excessive comments.
First point:
Yes, it is true that
$$\sum_{k=1}^n k\times\frac{k}{2}=1\times\frac{1}{2}+2\times\frac{2}{2}+...+n\times\frac{n}{2}$$
PROOF: Um... plug the numbers in. The way Sigma notation works is that
$$\sum_{k=m}^n f(k)=f(m)+f(m+1)+...+f(n-1)+f(n)$$
But we can extend this to non-indexed sets (as long as they are countable), e.g
$$\sum_{x\in\mathcal{X}}f(x)=f(x_1)+f(x_2)+f(x_3)+...~~\forall x_i\in\mathcal{X}.$$
Second Point:
Notice that I am calling this &quot;Sigma notation&quot; not a &quot;Sigma function&quot; as it really is just a way of writing things to make it shorter. You could, very formally, think of this instead as a map (&quot;function&quot;) that takes in two inputs - another function, and a set:
$$\Sigma(f,S)=f(s_1)+f(s_2)+f(s_3)+...~~\forall s_i\in S$$
E.g,
$$\Sigma(\text{id},\mathbb{N}_{\leq m})=\sum_{k=1}^m k=\frac{m(m+1)}{2}$$
But this is frankly overkill and unnecessarily complicates a simple idea.
Third point:
Yes, it is true that
$$\sum_{k=1}^n k\times\frac{k}{2}=\sum_{k=1}^n \frac{k\times k}{2}$$
There is nothing special about this! In a very crude manner, you can just &quot;cancel out&quot; the summation signs on both sides to get
$$k\times \frac{k}{2}=\frac{k\times k}{2}$$
Which is obviously true, e.g $(6\times 6)/2=36/2=18$ and $6\times (6/2)=6\times 3=18.$
Adding this as an answer to avoid excessive comments.
First point:
Yes, it is true that
$$\sum_{k=1}^n k\times\frac{k}{2}=1\times\frac{1}{2}+2\times\frac{2}{2}+...+n\times\frac{n}{2}$$
PROOF: Um... plug the numbers in. The way Sigma notation works is that
$$\sum_{k=m}^n f(k)=f(m)+f(m+1)+...+f(n-1)+f(n)$$
But we can extend this to non-indexed sets (as long as they are countable), e.g
$$\sum_{x\in\mathcal{X}}f(x)=f(x_1)+f(x_2)+f(x_3)+...~~\forall x_i\in\mathcal{X}.$$
Second Point:
Notice that I am calling this &quot;Sigma notation&quot; not a &quot;Sigma function&quot; as it really is just a way of writing things to make it shorter. You could, very formally, think of this instead as a map (&quot;function&quot;) that takes in two inputs - another function, and a set:
$$\Sigma(f,S)=f(s_1)+f(s_2)+f(s_3)+...~~\forall s_i\in S$$
E.g,
$$\Sigma(\text{id},\mathbb{N}_{\leq m})=\sum_{k=1}^m k=\frac{m(m+1)}{2}$$
But this is frankly overkill and unnecessarily complicates a simple idea.
Third point:
Yes, it is true that
$$\sum_{k=1}^n k\times\frac{k}{2}=\sum_{k=1}^n \frac{k\times k}{2}$$
There is nothing special about this! In a very crude manner, you can just &quot;cancel out&quot; the summation signs on both sides to get
$$k\times \frac{k}{2}=\frac{k\times k}{2}$$
Which is obviously true, e.g $(6\times 6)/2=36/2=18$ and $6\times (6/2)=6\times 3=18.$