Back when I was a tutor at the Bay Area Tutoring Center in San Ramon, I was often assigned to students that came in with calculus or pre-calculus concerns, leaving the middle school geometry and algebra to the regular turnover of younger tutors that starter with the easier courses. I had several students at this level, and the standard curriculum was just reaching the wonderful world of series, with students learning how to sum finite series and using the various rules and tests to determine whether a series converges (conditionally or not) or diverges. Most students found this chapter to be not-too-difficult, and were doing well with the assignments. One student was having difficulty with the material, so I asked him what type of series Grandi's Series was:
\[\sum_{k=0}^{\infty}(-1)^k=1-1+1-1+\cdots\]
Does it converge or diverge, how to prove it, and the like. After thinking about it for a minute, he asked "Well, doesn't it depend on whether infinity is odd or even?" A cute question; I gave him a wry smile and reminded him that infinity is not a real number, and that oddness or evenness can only apply to integers ("Is 1/7 odd or even?"). He nodded, then had a look of realization. "Oh, it's a geometric series, so we can use the formula."
\[\sum_{k=0}^{\infty}r^k=\frac{1}{1-r}.\]
"It must be convergent, since the answer is \(\tfrac{1}{2}\)." A gentle reminder about the radius of convergence of \(|r|<1\) and an appeal to the \(n^{\text{th}}\)-term test prompts another response "OK, so it diverges." Indeed it does; I mean, just look at the partial sums. It alternates between \(0\) and \(1\) forever, so how can it settle to a limit value? I demonstrate that such series are pathological by proving that \(0=1\):
\[\begin{align*}
0&=0+0+0+0+\cdots\\
&=(1-1)+(1-1)+\cdots\\
&=1+(-1+1)+(-1+1)+\cdots\\
&=1
\end{align*}\]
"The associative rule doesn't work for divergent infinite sums. In fact it only works for absolutely convergent sums." That smacks of the Riemann Rearrangement Theorem, but since that technically only applies to conditionally convergent series, I bite my tongue. "But doesn't it still average to 1/2?" I frown. The student has come to realize that like so many other divergent series that do occasionally show up in math and physics, some of them may diverge, but still WANT to be a particular value. That is, there are ways to unambiguously assign unique finite values to otherwise divergent sums. "For now, we only need to distinguish between divergent and convergent." He understands that it is divergent, and we move on.
That night, I think more about his question, and remember the ridiculous and yet useful fact that we may assign the sum of all natural numbers the value -1/12, using the peculiar properties of analytically continuing the Riemann zeta function. "What other such series might be useful? Should I enumerate them?" I decided to. Here is a list I've compiled:
\[\begin{equation*}
\begin{array}{>{\displaystyle}l@{}>{\displaystyle}l>{\displaystyle}l}
1-1+1-1+\cdots&{}=\sum_{k=0}^{\infty}(-1)^k&{}=\frac{1}{2}\\
1+1+1+1+\cdots&{}=\sum_{k=0}^{\infty}1^k&{}=-\frac{1}{2}\\
1+2+4+8+\cdots&{}=\sum_{k=0}^{\infty}2^k&{}=-1\\
1-2+4-8+\cdots&{}=\sum_{k=0}^{\infty}(-1)^k2^k&{}=\frac{1}{3}\\
1+z+z^2+z^3+\cdots&{}=\sum_{k=0}^{\infty}z^k&{}=\frac{1}{1-z}&\\
1+2+3+4+\cdots&{}=\sum_{k=0}^{\infty}k&{}=-\frac{1}{12}\\
1+4+9+16+\cdots&{}=\sum_{k=0}^{\infty}k^2&{}=0\\
1+2^n+3^n+4^n+\cdots&{}=\sum_{k=0}^{\infty}k^n&{}=-\frac{B_{n+1}}{n+1}\\
1-2+3-4+\cdots&{}=\sum_{k=0}^{\infty}(-1)^{k+1}k&{}=\frac{1}{4}\\
1+1+2+6+24+\cdots&{}=\sum_{k=0}^{\infty}k!&{}=?\\
1-1+2-6+24-\cdots&{}=\sum_{k=0}^{\infty}(-1)^kk!&{}=eE_1(1)\approx0.596\\
1-z+2z^2-6z^3+\cdots&{}=\sum_{k=0}^{\infty}(-1)^kk!z^k&{}=\frac{e^{1/z}}{z}\Gamma(0,1/z)\\
1+e^{i\theta}+e^{2i\theta}+\cdots&{}=\sum_{k=0}^{\infty}e^{ik\theta}&{}=\frac{1}{2}\left(1+i\cot\frac{\theta}{2}\right)\\
1+\cos\theta+\cos2\theta+\cdots&{}=\sum_{k=0}^{\infty}\cos
k\theta&{}=\frac{1}{2}\\
\sin\theta+\sin2\theta+\cdots&{}=\sum_{k=0}^{\infty}\sin k\theta&{}=\frac{1}{2}\cot\frac{\theta}{2}\\
1-\cos\theta+\cos2\theta-\cdots&{}=\sum_{k=0}^{\infty}(-1)^k\cos
k\theta&{}=\frac{1}{2}\\
\sin\theta-\sin2\theta+\cdots&{}=\sum_{k=0}^{\infty}(-1)^k\sin k\theta&{}=\frac{1}{2}\tan\frac{\theta}{2}\\
1^{2n}-2^{2n}+3^{2n}-\cdots&{}=\sum_{k=0}^{\infty}(-1)^kk^{2n}&{}=0\\
1^{2n+1}-2^{2n+1}+3^{2n+1}-\cdots&{}=\sum_{k=0}^{\infty}(-1)^kk^{2n+1}&{}=(-1)^n\frac{2^{2n+2}-1}{2n+2}B_{n+1}\\
1^{2n+1}-3^{2n+1}+5^{2n+1}-\cdots&{}=\sum_{k=0}^{\infty}(-1)^k(2k+1)^{2n+1}&{}=0\\
e^{i\theta}-e^{3i\theta}+e^{5i\theta}-\cdots&{}=\sum_{k=0}^{\infty}(-1)^ke^{(2k+1)i\theta}&{}=\frac{1}{2}\sec\theta\\
1^{2n}-3^{2n}+5^{2n}-\cdots&{}=\sum_{k=0}^{\infty}(-1)^k(2k+1)^{2n}&{}=\frac{(-1)^n}{2}E_n\\
\ln 2+\ln 3+\ln 4+\cdots&{}=\sum_{k=1}^{\infty}\ln k&{}=\ln\sqrt{2\pi}\\
2\ln 2+3\ln 3+4\ln 4+\cdots&{}=\sum_{k=1}^{\infty}k\ln k&{}=\ln A-\frac{1}{12}\approx 0.1654
\end{array}
\end{equation*}\]
Some of these come from extending the geometric series to all \(r\neq 1\), some from zeta function continuation, some from integral transforms. Some involve Bernoulli numbers and Euler numbers, others exponential integrals or the incomplete gamma function. This craziness can even be used to show that
\[1\times2\times3\times4\times\cdots=\prod_{k=1}^{\infty}k=\sqrt{2\pi}.\]
Perhaps asking about the parity of infinity is not so crazy after all.
\[\sum_{k=0}^{\infty}(-1)^k=1-1+1-1+\cdots\]
Does it converge or diverge, how to prove it, and the like. After thinking about it for a minute, he asked "Well, doesn't it depend on whether infinity is odd or even?" A cute question; I gave him a wry smile and reminded him that infinity is not a real number, and that oddness or evenness can only apply to integers ("Is 1/7 odd or even?"). He nodded, then had a look of realization. "Oh, it's a geometric series, so we can use the formula."
\[\sum_{k=0}^{\infty}r^k=\frac{1}{1-r}.\]
"It must be convergent, since the answer is \(\tfrac{1}{2}\)." A gentle reminder about the radius of convergence of \(|r|<1\) and an appeal to the \(n^{\text{th}}\)-term test prompts another response "OK, so it diverges." Indeed it does; I mean, just look at the partial sums. It alternates between \(0\) and \(1\) forever, so how can it settle to a limit value? I demonstrate that such series are pathological by proving that \(0=1\):
\[\begin{align*}
0&=0+0+0+0+\cdots\\
&=(1-1)+(1-1)+\cdots\\
&=1+(-1+1)+(-1+1)+\cdots\\
&=1
\end{align*}\]
"The associative rule doesn't work for divergent infinite sums. In fact it only works for absolutely convergent sums." That smacks of the Riemann Rearrangement Theorem, but since that technically only applies to conditionally convergent series, I bite my tongue. "But doesn't it still average to 1/2?" I frown. The student has come to realize that like so many other divergent series that do occasionally show up in math and physics, some of them may diverge, but still WANT to be a particular value. That is, there are ways to unambiguously assign unique finite values to otherwise divergent sums. "For now, we only need to distinguish between divergent and convergent." He understands that it is divergent, and we move on.
That night, I think more about his question, and remember the ridiculous and yet useful fact that we may assign the sum of all natural numbers the value -1/12, using the peculiar properties of analytically continuing the Riemann zeta function. "What other such series might be useful? Should I enumerate them?" I decided to. Here is a list I've compiled:
\[\begin{equation*}
\begin{array}{>{\displaystyle}l@{}>{\displaystyle}l>{\displaystyle}l}
1-1+1-1+\cdots&{}=\sum_{k=0}^{\infty}(-1)^k&{}=\frac{1}{2}\\
1+1+1+1+\cdots&{}=\sum_{k=0}^{\infty}1^k&{}=-\frac{1}{2}\\
1+2+4+8+\cdots&{}=\sum_{k=0}^{\infty}2^k&{}=-1\\
1-2+4-8+\cdots&{}=\sum_{k=0}^{\infty}(-1)^k2^k&{}=\frac{1}{3}\\
1+z+z^2+z^3+\cdots&{}=\sum_{k=0}^{\infty}z^k&{}=\frac{1}{1-z}&\\
1+2+3+4+\cdots&{}=\sum_{k=0}^{\infty}k&{}=-\frac{1}{12}\\
1+4+9+16+\cdots&{}=\sum_{k=0}^{\infty}k^2&{}=0\\
1+2^n+3^n+4^n+\cdots&{}=\sum_{k=0}^{\infty}k^n&{}=-\frac{B_{n+1}}{n+1}\\
1-2+3-4+\cdots&{}=\sum_{k=0}^{\infty}(-1)^{k+1}k&{}=\frac{1}{4}\\
1+1+2+6+24+\cdots&{}=\sum_{k=0}^{\infty}k!&{}=?\\
1-1+2-6+24-\cdots&{}=\sum_{k=0}^{\infty}(-1)^kk!&{}=eE_1(1)\approx0.596\\
1-z+2z^2-6z^3+\cdots&{}=\sum_{k=0}^{\infty}(-1)^kk!z^k&{}=\frac{e^{1/z}}{z}\Gamma(0,1/z)\\
1+e^{i\theta}+e^{2i\theta}+\cdots&{}=\sum_{k=0}^{\infty}e^{ik\theta}&{}=\frac{1}{2}\left(1+i\cot\frac{\theta}{2}\right)\\
1+\cos\theta+\cos2\theta+\cdots&{}=\sum_{k=0}^{\infty}\cos
k\theta&{}=\frac{1}{2}\\
\sin\theta+\sin2\theta+\cdots&{}=\sum_{k=0}^{\infty}\sin k\theta&{}=\frac{1}{2}\cot\frac{\theta}{2}\\
1-\cos\theta+\cos2\theta-\cdots&{}=\sum_{k=0}^{\infty}(-1)^k\cos
k\theta&{}=\frac{1}{2}\\
\sin\theta-\sin2\theta+\cdots&{}=\sum_{k=0}^{\infty}(-1)^k\sin k\theta&{}=\frac{1}{2}\tan\frac{\theta}{2}\\
1^{2n}-2^{2n}+3^{2n}-\cdots&{}=\sum_{k=0}^{\infty}(-1)^kk^{2n}&{}=0\\
1^{2n+1}-2^{2n+1}+3^{2n+1}-\cdots&{}=\sum_{k=0}^{\infty}(-1)^kk^{2n+1}&{}=(-1)^n\frac{2^{2n+2}-1}{2n+2}B_{n+1}\\
1^{2n+1}-3^{2n+1}+5^{2n+1}-\cdots&{}=\sum_{k=0}^{\infty}(-1)^k(2k+1)^{2n+1}&{}=0\\
e^{i\theta}-e^{3i\theta}+e^{5i\theta}-\cdots&{}=\sum_{k=0}^{\infty}(-1)^ke^{(2k+1)i\theta}&{}=\frac{1}{2}\sec\theta\\
1^{2n}-3^{2n}+5^{2n}-\cdots&{}=\sum_{k=0}^{\infty}(-1)^k(2k+1)^{2n}&{}=\frac{(-1)^n}{2}E_n\\
\ln 2+\ln 3+\ln 4+\cdots&{}=\sum_{k=1}^{\infty}\ln k&{}=\ln\sqrt{2\pi}\\
2\ln 2+3\ln 3+4\ln 4+\cdots&{}=\sum_{k=1}^{\infty}k\ln k&{}=\ln A-\frac{1}{12}\approx 0.1654
\end{array}
\end{equation*}\]
Some of these come from extending the geometric series to all \(r\neq 1\), some from zeta function continuation, some from integral transforms. Some involve Bernoulli numbers and Euler numbers, others exponential integrals or the incomplete gamma function. This craziness can even be used to show that
\[1\times2\times3\times4\times\cdots=\prod_{k=1}^{\infty}k=\sqrt{2\pi}.\]
Perhaps asking about the parity of infinity is not so crazy after all.