Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
I've tried to correct the link. I hope that the result of this edit is what you originally intended. –  Martin Sleziak Oct 26 '12 at 17:12
    
This is really a case of "why would you think you could make this sort of change in variables?" The only reason I can see is that you'd be working by analogy from integrals, but it doesn't even pass a simple smell test, as you've shown with finite examples. –  Thomas Andrews Oct 26 '12 at 17:12

4 Answers 4

up vote 4 down vote accepted

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.

share|improve this answer
    
Or one can write $$\sum_{n=0\atop n\text{ even}}^\infty$$ –  Hagen von Eitzen Oct 26 '12 at 17:22

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.

share|improve this answer

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 $$\sum_{k=a}^\infty f(k) = f(a)+\sum_{k=a+1}^\infty f(k).$$

share|improve this answer

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:

$$\sum_{n=1}^\infty\frac{1}{2^n}=\frac{1}{1-\frac{1}{2}}=2\neq \frac{4}{3}=\frac{1}{1-\frac{1}{4}}=\sum_{n=0}^\infty\frac{1}{4^n}=\sum_{n=0}^\infty\frac{1}{(2^2)^n}=\sum_{n=0}^\infty\frac{1}{2^{2n}}$$

share|improve this answer
    
Yes, coming up with example wasn't the hard part, but somehow I could get a intuitive reason of why it failed –  Jean-Sébastien Oct 26 '12 at 17:29

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.