Calculate
$$\sum_{n = 0}^{+\infty} a_n,$$
where
$$a_n = \begin{cases} \frac{1}{2^n} & n\ \text{even} \\\frac{1}{3^n} & n\ \text{odd} \end{cases}$$
What I thought: I split the series into even and odd series, and then I just use geometric series but I divide by two each term, since summin on only odd (or even) numbers, halves the sum.
Is that right?
I mean for example
$$\sum_{n\ \text{odd}}^{+\infty} \frac{1}{3^n} = \frac{1}{2}\sum_{n = 0}^{+\infty} \frac{1}{3^n}$$
Right?
No, that is not correct. Would you do that if you had only four terms? That is, do you believe that$$\sum_{n\in\{1,3\}}\frac1{3^n}=\frac12\sum_{n=0}^3\frac1{3^n}?$$I hope not, since the number on the left is $\frac13+\frac1{3^3}$, whereas the number on the right is $\frac12\left(\frac13+\frac1{3^2}+\frac1{3^3}+\frac1{3^4}\right)$.
In fact\begin{align}\sum_{n\text{ odd}}\frac1{3^n}&=\frac13+\frac1{3^3}+\frac1{3^5}+\cdots\\&=\frac13\left(1+\frac19+\frac1{9^2}+\cdots\right)\\&=\frac13\times\frac1{1-\frac19}\\&=\frac38,\end{align}whereas$$\frac12\sum_{n=0}^\infty\frac1{3^n}=\frac34.$$