Summation -infinity to +infinity

[johnbobson]

Hello. Please solve the following problem step by step.

∑[n=-∞,∞] 4

I know the answer is ∞.

The way I thought of solving it was to use ∑[n=m,n] 1 = [n+ 1 - m] so that would yield
∑[n=-∞,∞] 4 = 4[∞+ 1 - (-∞)]= 4[∞+ 1 +∞]= 4[2∞ + 1 ]= 4[2∞]= 8∞= ∞

Is this the correct way to work out he problem?

I have no experience with summations and I was wondering what category of math they fall into? I hope I am posting to the correct section.

 

[lookagain]

∑[n=-∞,∞] 4I know the answer is ∞. > > > Is this the correct way to work out he problem? < < <

johnbobson, it is allegedly *a* correct way. <------ Edit


Notice that each successive term isn't smaller than the previous term, and that the limit of \(\displaystyle a_n \) as n \(\displaystyle \to \infty \) is not 0 (where \(\displaystyle a_n = 4\).)

 

[HallsofIvy]

Hello. Please solve the following problem step by step.

∑[n=-∞,∞] 4

I know the answer is ∞.

The way I thought of solving it was to use ∑[n=m,n] 1 = [n+ 1 - m] so that would yield
∑[n=-∞,∞] 4 = 4[∞+ 1 - (-∞)]= 4[∞+ 1 +∞]= 4[2∞ + 1 ]= 4[2∞]= 8∞= ∞

Is this the correct way to work out he problem?

I have no experience with summations and I was wondering what category of math they fall into? I hope I am posting to the correct section.

I would say "no, that is not correct", although it can be used as "shorthand". What you saying is that in summing 4 from -m to n, you have 4(m+ n+ 1) and that gets larger without bound as m and n increase so the limit does not exist. Specifically, it is NOT correct to do arithmetic with "\(\displaystyle \infty\)".