# Sum of $1+2+4+8+…$ [duplicate]

I was solving a recurrence problem which had a sequence such as $y = (1+2+4+8+...)\sqrt n$, and I wanted to find what $x = 1+2+4+8+...$ was. So consider $x = 1+2+4+8+...$ as an infinite series.

$$x-1 = 2+4+8+...$$

$$2(x)=2(1+2+4+8+...) = 2+4+8+... = x-1$$

$$2x = x-1 \Rightarrow x=-1$$

What does $x=-1$ represent? Have I made a mistake in my calculations? Of course, we expect that $x=\infty$ since the sum grows by a factor of $2$ with each term.

## marked as duplicate by Jonas Meyer, Eric Stucky, Hans Lundmark, Hakim, 5xumOct 22 '14 at 8:41

• @JonasMeyer Didn't see that one. Thanks! – Bob Jonas Oct 22 '14 at 6:47
• You must formalize the sums with the limits and after you do all calculus taking the limit when $n\to \infty$. You cant directly manipulate infinity amounts, just finite sums. – Masacroso Oct 22 '14 at 8:38

By the standard definition, the value $$\sum_{n=1}^\infty a_n = a_1 + a_2 + a_3 + \cdots$$

is equal to the limit $$\lim_{N\to\infty}\sum_{n=1}^Na_n$$ IF this limit exists, and is not defined if the limit does not exist. In your case, the limit does not exist, therefore the sum is not defined, therefore marking the sum $x$ does NOT magically allow you to treat it as a number. The equation $x-1=2x$ simply isn't true.

There are some (for undergraduates, let's call them "non-standard") mathematical tricks in which you can assign some sort of number to infinite divergent sums using tricks like the one you used in your question. For that, see this article on Ramanujan sumation.

Important: Note that I did not say "summing up the divergent sums", but rather "assigning a number to the divergent sum", because, as I already explained, in the standard definition of the sum of an infinite series, divergent sums cannot be summed.

• $x-1=2x$ isn't too troubling if $x=$"$\infty$", but $2x-x = x$ is. – Jonas Meyer Oct 22 '14 at 6:41

This a case of not classical convergence and it makes perfect sense in some contexts : http://en.wikipedia.org/wiki/1_%2B_2_%2B_4_%2B_8_%2B_%E2%8B%AF

Since $x$ is not a constant number, in fact, it tends $\infty$, $\pm$ to $x$ is nonsense.

So you cannot have $2x=x-1$

• I am afraid this misses the target since, actually, $x=+\infty$ does solve $2x=x-1$. – Did Oct 28 '14 at 8:14

When you are writing "Let's call $x = 1+2+4+...$. It verifies...", what you are really saying is "If the limit $1+2+4+...$ exists, then it would verify...".

Since the limit of $1+2+4+...$ is $+\infty$, what you conclude from your equation isn't interesting.

Here you manipulate x as if it is finite numbers, so you have $2x$. But in fact $n\cdot\infty = \infty$ for any finite number $n$. And also $n+\infty=\infty$ again for any finite number $n$. So basically your equation is $\infty=\infty$, which have nothing useful.

You might be interested in this proof which shows that:

$1+2+3+...=-1/12$

The proof set $c=1+2+3+...$ and have $4c=4+8+12+...$, then he got $c-4c = 1-2+3-4+...=1/(1+1)^2$, ans so $c=-1/4$, which is absurd.

As said in wikipedia I linked to,

It is dangerous to manipulate infinite series as if they were finite sums.

• While this link may answer the question, it is better to include the essential parts of the answer here and provide the link for reference. Link-only answers can become invalid if the linked page changes. – Shaun Oct 22 '14 at 8:48
• Thanks for your suggestion. @Shaun – taper Oct 22 '14 at 8:57