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I'm getting two different answers for the following series...

$$\sum_{k=0}^{\infty} q^{2k}$$

With $-1 < q < 1$, I can solve the series by using the infinite geometric sum formula.

$$\sum_{k=0}^{\infty} q^{2k} = \frac{1}{1-q^2}$$

However, if I make the substitution, $m=2k$, then I get a different result.

$$\sum_{k=0}^{\infty} q^{2k} = \sum_{m=0}^{\infty} q^{m} = \frac{1}{1-q}$$

Where am I going wrong?

Yulmart
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    Compare... is $1+x^2+x^4+x^6$ the same sum as $1+x+x^2+x^3+x^4+x^5+x^6$? If you only use the even terms, is that the same as using all terms? – JMoravitz Oct 14 '18 at 20:26

2 Answers2

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By making the substitution $m=2k$ you are creating a sum that adds all the powers of $q$, not just the even powers, which is what your original sum does. For example, $q^3$ appears in the right-hand sum, but not in the original sum.

jeff honky
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What you were doing with the substitution $m = 2k$ was resulting in $$\sum_{m=0}^∞ q^m = q^0+q^1+q^2+q^3+q^4...$$ but the original equation yields $$\sum_{k=0}^∞ q^{2k} = q^0+q^2+q^4...$$ leaving a difference of $$\sum_{k=0}^∞ q^{2k+1} = q^1+q^3+q^5...$$ Instead of $m = 2k$, have $m = q^2$. Then, from $-1 < q < 1$ we get $0 \leq m < 1$. Then, $$\sum_{k=0}^∞ (q^2)^k = \sum_{k=0}^∞ (m)^k$$ and from this, because of the domain of m, we can apply the sum of an infinite geometric series formula. $$\frac{1}{1-m}=\frac{1}{1-q^2}$$

Joseph Gatts
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