What is the Proof that 2 Equals 1 Using Infinite Series?

[gnome]

I don't think we are entitled to invent our own meaning for a summation.

Summation has a very precise meaning which you can look up. My understanding of it is:
∑_k=1^m f(k) means you must specify some function of k (which may be just k itself, or k², or sin(k), etc), and you evaluate that f(k) at each of the integer values from k=1 to k=m (and hence you must specify what m is) and then add up all of those values of f(k).

That doesn't appear to be what you are doing.

 

[STAii]

Well, the function after the sum does not necessarily have to be a function of k (the variable that the sum is changing). Examples include :
∑(from k=1 to 3)1
which equals :
1+1+1=3
Or even, suppose that in your example, f(k) was a constant function, what will you have ? (well, you will have sth like what is above).
Mainly, if the function after the sum is not a function of the changing variable (k in our case), the value of the sum becomes as follows :
∑(from k=n to m)f(x₁)=(m-n+1)f(x₁)

And secondly, the value of m in my sum is specified, it has the value of [ x ].

Anything else

 

[quartodeciman]

Even a joke can have more than one use.

More fun!

The rightside function is indeed an infinite series of functions. Just look at the known values:

f(0) = 0 + 0 + 0 + 0 + ...
f(1) = 1 + 0 + 0 + 0 + ...
f(2) = 2 + 2 + 0 + 0 + ...
f(3) = 3 + 3 + 3 + 0 + ...
f(4) = 4 + 4 + 4 + 4 + ...

and so on.

So, how does one write this function in a closed form as an infinite series (paying attention to the values between those positive integer argument values)?

Enter the positive unit stepup function S_n(x). It is equal to 0 when x < n and it is equal to 1 when n<=x.

-S_n(x) is equal to 0 when x < n and it is equal to -1 when n<=x.

1-S_n(x) is equal to 1 when x < n and it is equal to 0 when n<=x. So, 1-S_n(x) is a positive unit stepdown function.

Check this out.
S_n(x) (1-S_n+1(x)) is equal to 0(0) = 0 when x < n; it is equal to 1(1) = 1 when n<=x<n+1;it is equal to 1(0) = 0 when n+1<=x.
This is a positive unit impulse function between two integer values.

Using these, one can construct the terms of the infinite series for all integer argument values, including also the negative ones:

f(-0) = 0 + 0 + 0 + 0 + ...
f(-1) = 1 + 0 + 0 + 0 + ...
f(-2) = 2 + 2 + 0 + 0 + ...
f(-3) = 3 + 3 + 3 + 0 + ...
f(-4) = 4 + 4 + 4 + 4 + ...

and so on.

Here is my expression for the infinite series function:

f(x) = &Sigma;_n=0^&infin; {n(1-S_-n(x) + S_n(x)) + (x²-n²)((1-S_-n(x))(S_-n-1(x))+ ((S_n(x))(1-S_n+1(x))}

I assert that this series will reproduce the values at integer arguments, both negative and positive, just like the above examples.
Alas, the infinite series converges pointwise, but not uniformly. So differentiation term by term isn't justified.
Finally, despite the hopeless appearance of discontinuity and undifferentiability at those integer argument values, I assert that f(x) = x² for all real x.