Index change of variable with product This is probably a dumb question, but oh well here it is:
One day a student came to me and asked: Why isn't 
$$
\sum_{k=0}^{\infty} \frac{1}{2^{2k}}=\sum_{n=0}^{\infty} \frac{1}{2^{n}}$$
using a substitution $n=2k$ makes $n$ range from $0$ to infinity as well and so the change of variables should be correct. (We had just used change of variables of the form $n=k-1$.
Now I convinced him that you could not do such change of variables by looking at other examples, such as
$$
\sum_{k=1}^3 4k=4+8+12\neq \sum_{n=4}^{12}n,
$$
using $n=4k$ substitution. He seemed satisfied but I wasn't really. For some reason I could not find an intuitive way to explain this, where it should be simple. 
Seeing this question made me think about another example and made me realize I still haven't found a intuitive way for this.
 A: The problem with letting $n=2k$ and then claiming that $$\sum_{k=0}^{\infty} \frac{1}{2^{2k}}=\sum_{n=0}^{\infty} \frac{1}{2^{n}}\tag{1}$$ is that the summation notation presupposes that the index variable increments by $1$’s. If the $n$ could somehow carry with it the information that it runs only over even integers, there’s be no problem, but it can’t.
To put it a little differently, $\sum\limits_{k=a}^bf(k)$ implicitly specifies $\{a,a+1,a+2,\dots,b\}$ as the domain of the index variable; the notation in this form does not allow any other kind of domain. Other forms of summation notation do allow it; for example, we can write $$\sum_{k\in\Bbb N}\frac1{2^{2k}}=\sum_{n\in 2\Bbb N}\frac1{2^n}\;,$$ encoding the necessary information in the specified domain of the index variable.
A: A change of variable should be a bijective map. The map $n \mapsto 4n$ is not bijective on $\mathbb{N}$, and that's why we lose terms in discrete sums like those you mention.
A: You're already answered in general, yet in your case it can be made even easier using the well-known sum of a geometric series:
$$\sum_{n=1}^\infty\frac{1}{2^n}=\frac{1}{1-\frac{1}{2}}=2\neq \frac{4}{3}=\frac{1}{1-\frac{1}{4}}=\sum_{n=0}^\infty\frac{1}{4^n}=\sum_{n=0}^\infty\frac{1}{(2^2)^n}=\sum_{n=0}^\infty\frac{1}{2^{2n}}$$
A: Using $n=k-1$ you can establish
$$\tag1\sum_{k=a}^\infty f(k-1) = \sum_{n=a-1}^\infty f(n).$$
So it seems one can use $n=g(k)$ to establish
$$\tag2\sum_{k=a}^\infty f(g(k)) = \sum_{n=g(a)}^\infty f(n)$$
but there is no such general theorem.
On can show the validity of  (2) under certain conditions: 


*

*$g$ must be a bijection $\{x\in\mathbb N\mid x\ge a\}\to \{x\in\mathbb N\mid x\ge g(a)\}$

*$g$ must be eventually monotonuous.


For the first point, observer that injectivity is needed to avoid somesummands being added repeatedly and surjectivity to avoid some summands are left out.
The second condition can be relaxed if the series is abslutely convergent.
By the way the proof that the above conditions on $g$ make (2) valid can be made by induction using (1) and
$$\sum_{k=a}^\infty f(k) = f(a)+\sum_{k=a+1}^\infty f(k).$$
