Convergence/divergence of geometric series when $k = 1$?

Question

Usually, when we try to determine convergence/divergence, we simply find the quotient $k$, and if $|k|<1$ we say the series is convergent. At least this is how the textbooks present it.

In set notation, this is the same as saying $k \in (-1, 1)$, but I don't understand why this set has to be open.

I can understand that $k = -1$ wouldn't be seen as convergent, as it would change sign every time. But shouldn't $k = 1$ be considered to yield convergence? Obviously, finding the $n$th term would be a trivial matter, but at least it would be known.

Should we not say that convergence is found when $k \in (-1, 1]$?

Answer 1

The geometric sequence $\{k^n\}$ converges for $k \in (-1,1]$. Namely for $k \in (-1,1)$ it converges to $0$ and for $k = 1$ it converges to $1$.

The geometric series $\sum_{n=0}^\infty k^n$ converges for $k \in (-1,1)$ (to $\frac{1}{1-k}$) only.

For example, for $k=1$ we have $\sum_{j=0}^n 1^j = \sum_{j=0}^n 1 = n+1$ which diverges to $\infty$ as $n \rightarrow \infty$.

Answer 2

When $k=1$, the geometric series

$$\sum_{n=1}^\infty k^n$$

does not converge, because it is equal to $$\sum_{n=0}^\infty 1$$

Meaning its partial sum is $$S_N=\sum_{n=1}^Nk^n=N$$

and the limit $$\lim_{N\to\infty} S_N$$ does not exist.