# Infinite amount of additions, finite sum?

I suggest it's a popular question, so if it was asked already (I couldn't find it anyway), close this question instead of downvoting, thanks!

Let's check this addition: $\sum_{n=0}^{\infty}\frac{1}{2^n}=2$

It looks like $1 + \frac12 + \frac14 + \frac18 + \frac1{16} + ...$ for every natural $n$.

How can the sum of this (and other similar) additions be finite? I tried to understand the concept of infinite, and if a variable is finite, then it always can be bigger than a value we give for it.

In the case of the example, by each time when I increase $n$ (I make it closer to infinite), I always add a number for the sum. The bigger $n$ is, the smaller this number is, so the sum always increases in a slighty smaller amount.

I got this fact as an answer even at the university. But I still don't understand it...$n$ always increases, so the sum is also always increases! Infinite is infinite, it can be always a bigger value, no matter what we write to $n$. For an incredibly big calue, the sum must be bigger. I mean, we can prove that for a given $n$, the sum becomes bigger than 1.7, or 1.8, kor 1.95, et cetera.

For an incredibly big $n$ value, even if its bigger than that we can even display (googolplex, for example), the sum should be bigger than 2.

...at least, theoretically. I don't get it. I've heard some "real-life" examples like eating $\frac1n$ table of chocolate every day, or the ancient Greek story (maybe from Archimedes) about the race of the turtle (or whatever), but it didn't help in understanding.

Is there anyone who can explain it on an understandable way?

• The sum of the first $n$ reciprocals of natural numbers is never $2$, and the harmonic series (where you take the limit of the sum as $n \to \infty$) diverges. Commented Sep 10, 2013 at 20:41
• $\sum_1^n \frac 1n = 2$ is wrong. (Even $\sum_{i=1}^n \frac 1i = 2$ is wrong for any $n$.) Commented Sep 10, 2013 at 20:41
• $\sum_1^n\frac1n=\frac1n+\frac1n+\dots+\frac1n=1$, this is not what you meant. On the other hand, $\sum_1^\infty\frac1n\ne 2$, but $\sum_0^\infty \frac1{2^n}=2$. Commented Sep 10, 2013 at 20:41
• Perhaps you meant $\sum_{n=0}^\infty\frac{1}{2^n}$? That is equal to $2$, as the partial sums are of the form $1, 1\frac{1}{2}, 1\frac{3}{4}, 1\frac{7}{8}, 1\frac{15}{16}$, etc. Note how each partial sum is bigger than the one before it, but they never get bigger than 2. Commented Sep 10, 2013 at 20:43
• Maybe you meant $\sum_{n=0}^{\infty}\frac{1}{2^n}=2$ Commented Sep 10, 2013 at 20:44

Consider a cake of size 2 (in whatever units). First day you eat half of it, that is, 1. Second day you eat half of what remains: 1/2. Third day half of what remains: 1/4. And so on for infinitely many days. How much will you eat in total? Clearly not more than 2, yet that finite sum is obtained from infinitely many summands.

So, if it confuses you that infinitely many summands can give a finite sum, think of it the other way around: you start with a finite number and decompose it into infinitely many parts.

• Until the second part, I'd have had some doubts, but with the last 2 lines, this example is also clear. Thx! Commented Sep 10, 2013 at 21:56

A simplified version of one of Zeno's paradoxes is that the runner will never arrive at the end, because it has do run the first half, then half of what remains, then half of what remains, …

But we see runners arrive, maybe a good deal less than ten seconds for the 100 meter race. So there's something wrong in what Zeno was telling us. But this is not pure mathematics, it's physics and we don't want to mix the two planes, because this might lead to circular reasoning.

Rather, we can see that, given any positive rational number $\varepsilon$, we can find that one of the sums $$1+\frac{1}{2}+\frac{1}{2^2}+\dots+\frac{1}{2^n}$$ is between $2-\varepsilon$ and $2$ and that none of those sums exceeds $2$. It's quite easy, actually. Since $$\sum_{k=0}^n \frac{1}{2^k}=\frac{1-\dfrac{1}{2^{n+1}}}{1-\dfrac{1}{2}} =2\left(1-\dfrac{1}{2^{n+1}}\right)=2-\frac{1}{2^n}$$ we have just to solve for $n$ the inequality $$2-\frac{1}{2^n}>2-\varepsilon,$$ that is, $$\frac{1}{\varepsilon}<2^n$$ which is surely satisfied for a sufficiently large value of $n$: just expand $1/\varepsilon$ in its binary representation, which will have a finite number of digits in its integral part; take the number $100\dots0$ (in binary, of course) having as many zeros as the digits in the integral part of $1/\varepsilon$. The other inequality boils down to $$2-\frac{1}{2^n}<2$$ which is obvious.

Therefore, if any meaning at all can be given to the infinite series, it must be $2$. And we define that infinite sum to be $2$. Then we study the algebraic and analytic properties of infinite series (convergent or even divergent), seeing that these properties are consistent with that definition of sum (provided the foundation of mathematics is consistent, but this is a quite different problem).

People apply the analytic concept to the real world, building mathematical models, and see that they describe the world pretty well. Or, in some cases, they don't: but then the model is wrong, not the math underlying it.

By using infinite series, Newton was able to do a darn good description of gravitation, without pretending that his mathematical theory was the ultimate truth about the physical world (he knew about some deep problems connected with his theory, which couldn't be “explained” easily, such as instantaneous interaction between bodies far away from each other). But, again, this is not strictly mathematics. As long as our mathematical definitions and reasonings are not contradictory, it's not a question of “believing” to the actual infinity: it's just seeing whether the proofs are right or wrong.

There's nothing contradictory in assigning the value $2$ to your series, so we do it. But we never do an infinite number of sums.

• Oh yes, that one of Zeno's paradoxes was I mentioned as an example I've known before. Also, +1 for the long answer and therefore the huge effort! Commented Sep 10, 2013 at 21:52

For an incredibly big $n$ value, even if its bigger than that we can even display (googolplex, for example), the sum should be bigger than $2$.

Well... no, for every $n$, $\sum\limits_{k=0}^n1/2^k\lt2$ hence no partial sum $\sum\limits_{k=0}^n1/2^k$ is greater than $2$ (and in fact none is equal to $2$).

To see this, note that $1/2^n+\sum\limits_{k=0}^n1/2^k=2$.

Here is an analogous situation, can you tell us if you find it mysterious as well:

Consider the series $\sum\limits_{k\geqslant0}9/10^k$. The partial sums are $9$, $9.9$, $9.99$, $9.999$, $9.9999$, and so on. Do you think that "for an incredibly big $n$ value, the sum should be bigger than $10$"?

• The example is not really important, tha matter is that there are infinite sums with finite value. But that latter example is perfectly understandable! The problem might have been that previous examples didn't highlight the fact that every addition adds a smaller number. Even school examples were terrible in this aspect - or at least, I missed this one. Thank you! Commented Sep 10, 2013 at 21:33

First of all note that $\displaystyle\sum_{i=1}^{\infty}\dfrac{1}{i} \neq 2$, it diverges. I'm not sure if this is they key to your misunderstanding, or if it goes beyond this particular series.

Let $(a_n)$ be a sequence of real numbers and write $S_k = \displaystyle\sum_{n=1}^ka_n$ for the sum of the first $k$ terms of the sequence; this is called a partial sum. Now note that it doesn't make sense to add infinitely many things together, only finitely many. Therefore we have to define what it means to add all the terms of the sequence $(a_n)$. We make the definition $\displaystyle\sum_{n=1}^{\infty}a_n = \lim_{k\to\infty}S_k = \lim_{k\to\infty}\sum_{n=1}^ka_n$. If the limit exists, we say that the series $\displaystyle\sum_{n=1}^{\infty}a_n$ converges, otherwise we say it diverges. Using the fact that $a_k = S_k - S_{k-1}$, one can show that if the series converges, then $\lim\limits_{k\to\infty}a_k = 0$. However, the converse is not true (consider the harmonic series $a_n = \frac{1}{n}$).

Back to your question. How can infinite sums be finite? Well an infinite sum is nothing but a limit, so you just need to convince yourself that there are sequences $(S_k)$ with finite limit.

Consider the example $a_n = 2^{-n}$. One can show that $S_k < 1$ for every $k$. However, $S_{k+1} = S_k + 2^{-(k+1)} > S_k$, so $(S_k)$ is an increasing sequence which is bounded above; it converges by the Monotone Convergence Theorem, in this case to $1$.

The key is that each term you add contributes less and less (this is what $\lim\limits_{k\to\infty}a_k = 0$ tells us). In the example above, after any number of finite terms, the sum is less than one, but it is still increasing. If it never surpasses one, the sum is never going to go to $\infty$.

• The example wasn't so important, the matter is that we make an infinite amount of additions. Anyway, I wouldn't call it a simplified explanation (:D), however I've learnt about it, so it' partially clear. I respect the effort of course. Commented Sep 10, 2013 at 21:24
• @ZoltánSchmidt: While I found your question interesting, I think at some point you just need to accept that adding an infinite amount of terms can result in a finite result, and the intuition will come later. Someone on this site once quoted something along the lines of "In math, we don't understand things; we just get used to them". Try to work out the $\sum \frac1{2^n}$ example. It's easy to do and it will give you some insight as to why the total sum is $2$ and not $\infty$. Commented Sep 10, 2013 at 21:30
• @ZoltanSchmidt: I'm not sure what you expect a simplified explanation to be. Without being precise you end up running into the traps that you have. The only way to escape these traps is to write out exactly what is meant and isolate the part(s) which you are unclear on. Commented Sep 10, 2013 at 21:35
• @MichaelAlbanese I got the answer from Did. The problem was that I couldn't see the exact meaning of "adding always a bit smaller number". Sorry if I couldn't explain it, my mistake. Commented Sep 10, 2013 at 21:38
• @JavierBadia Exactly accepting and understanding that fact was difficult for me. I'm not a mathematician, but for me, there's the most significant fact to understand what I use, to be able to use it right. At least, if there's a proof for it. Commented Sep 10, 2013 at 21:43

My sister asked me this today. I typed this up for her. Figured I may as well put it up here as well.

It can in the following sense:

What does it mean to say $$\lim_{n \to \infty} \ \frac{1}{n} = 0 \ ?$$

$\\$

Cleary $\frac{1}{n}$ is never zero. What it means is that as $n$ gets bigger and bigger and bigger, $\frac{1}{n}$ gets closer and closer to zero. We can make $\frac{1}{n}$ as close to $0$ as we want by making $n$ large enough.

$\\$

The definition for an infinite series is as follows:

$\\$

$$\sum_{k = 1}^\infty a_k \ := \lim_{n \to \infty} \sum_{k=1}^n a_k$$

Put another way, if we let $$s_n = \sum_{k=1}^n a_k$$ $\\$ $$\text{then } \ \sum_{k = 1}^\infty a_k = \lim_{n \to \infty} s_n$$

That it, it is the limit of the sequence $$s_1,s_2,s_3 \dots = a_1\ ,\ a_1 + a_2\ ,\ a_1+a_2+a_3, \ \dots$$

$\\$

Clearly we can't sum a number infinitely many times, and after any finite number of times, when we add the next term we get a bigger number then we had before. So if we keep summing wouldn't the numbers just continue getting bigger and bigger all the way to infinity? Well consider this sequence of numbers:

$\\$

$$.1,\ .11,\ .111,\ .1111,\ .11111,\ \dots$$

$\\$

Each number is strictly bigger than one before, but at no point does it ever go beyond $.12$ . These are actually the partial sums of the series $$\sum_{k=1}^\infty \frac{1}{10^k}$$

If the sequence $a_k$ converges to zero, then although we keep adding more and more numbers, the numbers we are adding get smaller and smaller, and so you see its possible for the resulting sums to not get arbitrarily big. Thinking back in terms of limits, when we say that $$e = \sum_{k = 0}^\infty \frac{1}{k!}$$

we mean that $e$ is the limit of the sequence $$\frac{1}{0!},\quad \frac{1}{0!}+ \frac{1}{1!}, \quad \frac{1}{0!}+ \frac{1}{1!}+ \frac{1}{2!}, \quad \dots$$

$\\$

In other words, as we keep adding more and more terms of the sequence, the sum gets closer and closer to $e$. We can get as close as we want to $e$ by adding enough terms.

It turns out that having $a_k$ converge to zero is not enough. It has to go to zero quickly. For instance the sequence $$\frac{1}{k}$$ converges to zero, but the series $$\sum_{k = 1}^\infty \frac{1}{k} = \infty$$

In this instance, as we keep adding more and more terms, the sums just get bigger and bigger. No matter what upper bound you try to bestow on it, eventually the sum will surpass it.

However, The sequence $$\frac{1}{k^2}$$ does shrink quickly enough. When I plug these terms into a computer I get the following for $\frac{1}{k} \text{ and } \frac{1}{k^2} \text{ respectively:}$

$$\begin{array}{c|c} n & \sum_{k = 1}^n \frac{1}{k} \\ \hline 1 & 1 \\ 2 & 1.5 \\ 3 & 1.8333 \\ 1000 & 7.48547 \\ 10,000 & 9.78761 \\ 100,000 & 12.0901 \\ 1,000,000 & 14.3927 \\ 2,000,000 & 15.0859 \end{array}$$

It grows slowly but it continues to grow nonetheless.

For $\frac{1}{k^2}$ I get:

$$\begin{array}{c|c} n & \sum_{k = 1}^n \frac{1}{k^2} \\ \hline 1 & 1 \\ 2 & 1.25 \\ 3 & 1.36111\\ 1000 & 1.64393\\ 10,000 & 1.64483 \\ 100,000& 1.64492\\ 1,000,000 & 1.64493 \\ 2,000,000 & 1.64493\end{array}$$

By the time that I got to one million, adding the next one million terms increased the sum by less than $0.00001$! That is, $$\sum_{k = 1,000,001}^{2,000,000} \frac{1}{k^2} < 0.00001$$

In this instance, The sequence $\frac{1}{k^2}$ shrinks quickly to ensure the resulting sequence of partial sums doesn't get arbitrarily big.

An infinite number of summands is not really a problem, if the summands decrease quickly enough. For example $$\lim_{n\to\infty} \sum_{k=1}^n \frac{1}{k^2} = \frac{\pi^2}{6}<\infty$$ But $$\lim_{n\to\infty} \sum_{k=1}^n \frac{1}{k} = \infty$$ So for the convergence of series (which is what you are puzzled about), there are a lot of so-called convergence tests for the summands $a_k$ to determine, whether $$\lim_{n\to\infty} s_n := \lim_{n\to\infty} \sum_{k=1}^n a_k$$ converges.

• But this is still not clear: "if the summands decrease quickly enough". As I mentioned, infinite means that it increases constantly, it can be bigger for every $k$ - either if I write $k$ or $k^2$ in the denominator, $1/k$ and $1/k^2$ never becomes zero, so sum always becomes a bit bigger if I add any of them to it. Commented Sep 10, 2013 at 20:57
• Quickly enough is, for example $$|a_k| \leq \frac{1}{k^\alpha}, \qquad \alpha > 1, k \geq K_0$$ (in general, the "start" of the sequence $(a_k)$ doesn't matter, only the behaviour "behind" a $K_0$) Commented Sep 10, 2013 at 20:59

The sum you have written is not correct. If instead you really did mean this:

$\sum_{n=0}^{\infty}\frac{1}{2^n}=2$

Then this can be visualised as follows:

Here the first filled-in circle represents the first term of the sum = 1
The second pair of circles represent the first two terms of the sum = 1 + 1/2
The third pair of circles represent the first three terms of the sum = 1 + 1/2 + 1/4
etc...

So hopefully here you can see how an infinite sum can lead to a finite value.

• I can [b]see[/b], this is not a problem. But it's against my sense, if infinite is [b]seriously[/b] infinite. Check my comment I wrote for the answer of AlexR. Commented Sep 10, 2013 at 21:00
• @ZoltánSchmidt -- the point is you should be able to see that each terms in this particular sequence keeps adding half of what is left in the second circle. So the total sum can never exceed 2. You just keep getting closer and closer to 2. In fact you will only get EXACTLY 2 when you have added an infinite amount of terms - otherwise, for any finite amount, you just keep getting closer and closer to 2. Commented Sep 10, 2013 at 21:02
• That's what I could barely understand. But it's clear now. Commented Sep 10, 2013 at 21:58