Proof of this trivial Ramanujan result

Question

The title is obviously sarcastic, and, sorry for my ignorance. Where can I find proofs for Ramanujan results like

$$1-1+1-1+1+...= \frac{1}{2}$$

$$1+2+3+4+5+...=-\frac{1}{12}$$

I don't seem to find anything around here... Thanks for the help!

Answer 1

For $1-1+1-1+1-\ldots =\frac{1}{2}$

Consider the sum $(1-1)+(1-1)+\ldots= 0+0+\ldots =0$

Now shift the parenthesis one number down so we have $1+(-1+1)+(-1+1)+\ldots=1+0+0+\ldots=1$

Taking the average of the two yields $\frac{1}{2}$.

Answer 2

Your first question has been asked before here:

Checking my understanding: $1 - 1 + 1 - 1 + 1 - ... = \frac{1}{2}$

$1−1+1−1+1−1+\cdots=\frac{1}{2}$ proof?

Your second question has an answer here:

Why does $1+2+3+\cdots = -\frac{1}{12}$?

There are also several similar questions on this site for example:

$1-2+3-4+\dots = \frac{1}{4}$

Series of logarithms $\sum_{k=1}^\infty\ln(k)$ (Ramanujan summation?)

Is it possible to assign a value to the sum of primes?

Is my $1+1+1+1+1...=-\frac{1}{2}$ proof correct?

More aboute the methods used to extract a finite number from divergent series can be found for example in:

Analytic continuation -Easy explanation?

Zeta function zeros and analytic continuation