Explain $\sum_{n=0}^\infty 2^n= 1+2+4+8+16+\cdots =-1$

Question

So I was playing around with, $$\sum_{n=0}^\infty 2^n= 1+2+4+8+16+... $$ Let $S$ be the sum $$S= 1+2+4+8+16+...$$ Dividing both sides by $2$ $$\frac{S}{2}=\frac{1+2+4+8+16+...}{2}$$ Which means $$\frac{S}{2}=\frac{1}{2}+1+2+4+8+...$$ Or $$\frac{S}{2}=\frac{1}{2} + S$$ Therefore after solving for $S$ we get, $$S=-1$$ I believe I'm doing something terribly wrong here but I am convinced that the math checks out. Please help. I am in $10$-th so please bear with me if this is a stupid idea and tell me why it's wrong, cause I know it's wrong.

Answer 1

There are a lot of phony identities you can get with this kind of logic. When you write down $$ \frac{S}{2} =\frac{1}{2} +S$$ you are assuming $S$ is some number, when it's clear from the form of the sum that it diverges to infinity.

In fact, this doesn't quite disagree with the equation you wrote down. If you "plug in infinity" to both sides, you see you get $\infty = \infty,$ so in a loose sense, $\infty$ is a valid solution to that equation. The solution $S=-1$ you got by solving for $S$ is also a solution, but it's a spurious one.

It often happens when you're looking for "fixed points" you'll get an equation with more than one solution and additional analysis is needed to figure out which one is the actual limit the system goes to.

Edit

Some people been talking of 'partial sums' and 'truncation' so I figure I'll spell it out in a little more detail what's happening here.

The finite sum $$ S_n = \sum_{k=0}^n 2^k$$ is called the partial sum of the series and the value of the infinite series is the limit $\lim_{n\to \infty}S_n.$ If this limit exists as a number we say the series converges, otherwise it diverges. We can do a version of your algebra on the partial sums: $$ \frac{1}{2}S_n = \frac{1}{2} + 1 \ldots +2^{n-1} = \frac{1}{2} +S_{n-1}=\frac{1}{2} + S_n - 2^{n}$$ which you can solve to give $$ S_n = 2^{n+1} -1.$$ This is the real solution to the partial sums. And we can see that $2^{n+1}-1$ grows and grows as $n\to \infty,$ so $\lim_{n\to\infty} S_n = \infty$ and the series diverges.

Now, here's something interesting: we can generalize this to $$ S_n = \sum_{k=0}^n a^n$$ where $a$ is some number and then run the algebra again: $$ \frac{1}{a} S_n = \frac{1}{a} + S_{n-1}=\frac{1}{a} + S_n - a^n$$ which you can solve to give $$ S_n = \frac{a^{n+1}-1}{a-1}.$$ Now, notice that if $a$ is a small number, like $1/2,$ then $a^{n+1}$ goes to zero as $n\to \infty$ and you get $S_n = \frac{1}{1-a}.$ So we have, for instance, $\sum_{n=0}^\infty (1/2)^n = 2.$

Notice that your wrong conclusion $S=-1$ for $a=2$ is what you'd get plugging in $a=2$ into $\frac{1}{1-a}.$ What you've done is forget that giant $2^{n+1}$ in the numerator of the true formula for $S_n.$ When you set both $S_{n-1}$ and $S_n$ equal to $S$ in $$ \frac{1}{2}S_n = \frac{1}{2} + S_{n-1},$$ you are assuming that the difference between them goes to zero, which is not true here... the difference is $2^n.$

Answer 2

"Series' summation doesn't make sense!", this is somewhat true,

when you have a series it can or diverge or converge,

Converge means that there is no single value that the series goes to, for example the series $\sum_{i=1}^\infty \frac{1}{i^2}$ is converges to a single value, $\frac{\pi^2}{6}$.

Diverge means that the series does not converge, in this case we can see that the series $\sum_{n=0}^\infty 2^n$ is always increasing($a_i<a_{i+1}$), this means that there is no value that the series will goes to.

we know do actions like you just did only if the series converges. Like @spaceisdarkgreen wrote if you "plug in infinity" you still get $\infty=\infty$, but this doesn't make much sense because infinity is not a number, we can say that $\frac{S}{2} =\Theta(\frac{1}{2} +S),$ i.e. $\frac{S}{2}$ is bounded asymptotically by $\frac{1}{2} +S$.

something to keep in mind is that diverge series not always goes to infinity, for example $\sum_{i=0}^\infty(-1)^i=1-1+1-1+\cdots$, this series does not go to infinity but it also does not goes to any particular value, it "jumps" from $-1$ to $1$, so it disverges