A Whole Party of Infinite Friends
We’ve seen the startling result that if we add up an infinite number of one-over-powers-of-two, we get just \(2\):
\[1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\cdots\quad = 2\]
This perplexity should provoke our curiosity. If we can sometimes add up an infinite number of things and get a finite number, why does that happen? When does it happen? Who are these alien creatures, these infinite series? How do they behave? When do they converge (approach some finite number), and when do they diverge (not)?
Here are twenty more extraterrestrials for you to meet. What can you learn about them? Put more conceretely: can you figure out which of these series converge, and which diverge? For those that converge, can you figure out what they converge to? Try to figure these out without looking anything up, without giving away any spoilers, just using your mind! Getting the answer isn’t the main goal: if figuring out the convergence/divergence were our top priority, then flipping a coin would take us halfway there. Rather, the point is to think and reason through these problems and make an argument.
- \(\displaystyle \sum_{n=0}^{n=\infty} \frac{1}{2^n}\)
- \(\displaystyle \sum_{n=3}^{n=\infty} \frac{1}{2^n}\)
- \(\displaystyle \sum_{n=0}^{n=\infty} 2^{-n}\)
- \(\displaystyle \sum_{n=0}^{n=\infty} 2^{n}\)
- \(\displaystyle \sum_{n=0}^{n=\infty} (-1)^n\)
- \(\displaystyle \sum_{n=0}^{n=\infty} 3\)
- \(\displaystyle \sum_{n=0}^{n=\infty} \frac{1}{3}\)
- \(\displaystyle \sum_{n=0}^{n=\infty} \frac{1}{5^n}\)
- \(\displaystyle \sum_{n=1}^{n=\infty} \frac{1}{n}\)
- \(\displaystyle \sum_{n=0}^{n=\infty} \frac{1}{n!}\)
- \(\displaystyle \sum_{n=0}^{n=\infty} \frac{1}{n! + 5^n + 200}\)
- \(\displaystyle \sum_{n=1}^{n=\infty} \frac{1}{\sqrt{n}}\)
- \(\displaystyle \sum_{n=-4}^{n=\infty} \frac{1}{2^n}\)
- \(\displaystyle \sum_{n=0}^{n=\infty} \frac{12}{2^n}\)
- \(\displaystyle \sum_{n=0}^{n=\infty} \frac{1}{7.842^n}\)
- \(\displaystyle \sum_{n=0}^{n=\infty} \frac{1}{2^{n+3}}\)
- \(\displaystyle \sum_{n=0}^{n=\infty} 2^{n}\) revisited
- \(\displaystyle \sum_{n=0}^{n=\infty} \frac{1}{2^n} + \frac{1}{3^n}\)
- \(\displaystyle \sum_{n=0}^{n=\infty} \frac{3^n + 2}{4^n}\)
- \(\displaystyle \sum_{n=0}^{n=\infty} \frac{1}{2^{2n+3}}\)