Proving that a series is divergent?

Question

My question:

Is proving that a series sum gives weird result when assumed to be a real number sufficient to prove that a series is divergent?

For example it is a famous proof that $\sum_{n=1}^\infty n= -\frac{1}{12}$. It is obvious that the series is a diverging one and assuming it to be a real number is giving the wrong result.

So if a series gives a weird sum like a negative sum of all positive terms or a fractional term of all whole numbers if we assume it to be a real number can we safely conclude that the series is a diverging one?

Thanks.

Answer 1

Short answer: no.

“Weird” is a very vague concept. I suppose that when Euler proved that$$\sum_{n=1}^\infty\frac1{n^2}=\frac{\pi^2}6,$$several persons found that weird. It is true nonetheless.

And if somoene tells you that $\sum_{n=1}^\infty\frac1{n^2+n}=-1$, I suppose that you will find that weird. But the series $\sum_{n=1}^\infty\frac1{n^2+n}$ converges (and the sum is $1$).

Answer 2

I would say: yes. This is just a proof by contraddiction. What I think you want to say is: if we assume that the series is convergent and we obtain a false result, is the series divergent? Of course.

In this case, you expect the result of convergence to be positive. If you get $-1/12$ with same manipulations that are allowed because you are assuming the series to converge to a number, you are done with your proof.

As an example, consider the (increasing) sequence $x_{n+1}=x_n+1/x_n$,starting with $x_1=1$. Can this sequence be convergent?