Is the equal symbol in an infinite series misleading notation?

Question

The infinite series in this notation: $$\frac12+\frac14+\frac18+\dots = 1$$ is nothing more than the limit of the partial sums in: $$\sum_{n=1}^\infty\frac{1}{2^{n}}$$ The initial notation implies to beginner students that somehow we are "summing a number of non-ending terms" and they magically give the result of 1. Although that is not the case, since it does not make sense to 'add' (in the conventional sense) a non-ending sequence of numbers (as it never ends). (or have I misunderstood that?)

Is the usage of the equal symbol in the first notation misleading/wrong? If not, then what does the equal symbol really mean?

Edit: Note: I do, of course, agree with the result of the summation and I understand the result. I am just wondering about the appropriateness of the notation.

Answer 1

The notation $$ \sum_{n=1}^\infty\frac{1}{2^{n}} \tag1$$ should be completely explained in any calculus textbook. The notation $$ \frac12+\frac14+\frac18+... = 1 \tag2$$ is a dumbed-down version for those who do not know what $\sum$ means, those who have never studied calculus, or other special purposes. I agree that $(2)$ with no extra explanation can be misleading.

Answer 2

An infinite sum is simply shorthand for the limit of a convergent series. You're right in the sense that infinite addition alone is not well-defined. But the limit of a summed series is perfectly well-defined.

Let $(a_n)=(a_1,a_2,...) \subset \mathbb{R}$ be a real sequence. Its summation series is a sequence $(s_n)$ defined as $$s_n=\sum_{k=1}^n a_k.$$ Now, whether $(s_n)$ converges or not is a different story. When $(s_n)$ is convergent, then $$s^*=\sum_{k=1}^\infty a_k$$ is simply shorthand for $$s^* = \lim_{k \to \infty} s_k,$$ which is a well-defined notion of equation.