I would like to get a help about the next problem from my textbook:
Problem:
For which values of an $a \in \Bbb{R}$ the sum of a geometric series $2a + \sqrt{2a}+ a + \cdots$ equals 8? (first i wrote $\sqrt{2}a$, but it is $\sqrt{2a}$)
My solution:
First of all, i don't think that $2a$, $\sqrt{2a}$ and $a$, represent first three memebers of some geometric sequence. I think that this is a typo, and that it should be $a + \sqrt2a + 2a + \cdots$.
In this case we have a sequence of next numbers: $a, a\sqrt{2}, a(\sqrt{2})^2, a(\sqrt{2})^3, \ldots$ Here we have that common ratio is $q = \sqrt{2}$. From here we have that $$S_n = a \frac{1 - (\sqrt{2})^n}{1 - \sqrt{2}},$$ which gives us $$\lim_{n \to 0}S_n = \lim_{n \to \infty} a \frac{1 - (\sqrt{2})^n}{1 - \sqrt{2}} = \frac{a}{-(\sqrt{2} - 1)} \lim_{n \to \infty}(1 - (\sqrt{2})^n) = +\infty.$$ So, we have that this series converges to $+\infty$ and there isn't a $a \in \Bbb{R}$ for which the sum of given series is equal to 8 (the partial sums aren't the part of an answer).
Please, could you tell me if i made a mistake somewhere or the problem is in the typo and something else should be written?
I think it should be $$2a+\sqrt2a+a+...=8.$$
$$q=\frac{\sqrt{2}a}{2a}=\frac{1}{\sqrt{2}}.$$ Thus, $$\frac{2a}{1-\frac{1}{\sqrt{2}}}=8.$$ Can you end it now?