Is $\sum_{k=-\infty}^{0}f(-k)=\sum_{k=0}^{\infty} f(k)$ always?

Question

I wonder, whether it is always the case

$$\sum_{k=-\infty}^{0}f(-k)=\sum_{k=0}^{\infty} f(k)$$

in regards of summation methods for divergent series?

Answer 1

No. For example if $f(k):=e^k$, then one of the sums converges, the other diverges.

Answer 2

Yes. As for the proof, it's quite simple.

$s_n=f(0)+f(1)+\ldots+f(n)$

$s_{-n}=f(0)+f(-(-1))+f(-(-2))+\ldots+f(-(-n))=f(0)+f(1)+f(2)+\ldots+f(n)$

That is, $s_n=s_{-n}$

As $n\to-\infty$, so $-n\to+\infty$. That is, both series converge or diverge together and to the same value.

However, take note of how I defined the infinite series on the left, making both series analagous. Ask if you have any question.

Answer 3

Here is another "no".

One can work with "bilateral" summation methods for divergent series of the form $$ \sum_{k=-\infty}^\infty a_k $$ There is no reason such a method should be "symmetric" in the sense of your question. Nor, more generally, need the sums assigned to $$ \sum_{k=-\infty}^\infty a_k \qquad\text{and}\qquad \sum_{k=-\infty}^\infty a_{-k} $$ be the same. Some bilateral methods are symmetric, of course, but others are not.