$.99999...=1$: What is the proof that we can multiply an infinite series by something to shift it?

Question

I wanted to prove to myself that $.9999... = 1$ so I wanted to show that:

$N = 9/10 + 9/100 + 9/1000 + 9/100000 + ...$

or more formally, $N = 9S$ where

$S = \sum_{k=1}^{\infty} \frac{1}{10^k}$

Then multplying by $10$ I get

$10S = \sum_{k=1}^{\infty} \frac{1}{10^{k-1}} = \sum_{k=0}^{\infty} \frac{1}{10^{k}}$

Subtracting I get

$10S - S = \sum_{k=0}^{\infty} \frac{1}{10^k} - \sum_{k=1}^{\infty} \frac{1}{10^{k}} = \frac{1}{10^0} = 1$

So if $9S = 1$, then $S = 1/9$, and $N = 9S = 1$, proof completed.

1. Is this proof correct?

2. What is it exactly that allows me to multiply an infinite series by $10$ and have it be a valid transformation? To my "ignorant intuition" this seems like taking something infinite and making it $10$ times infinity, which I don't know what that really means.

3. What is it that allows me to shift indexes like I did to move $k$ down from $1$ to $0$? Similar to question 2 above. We shift and yet do not have to touch or manipulate the upper bound of infinity at all. I know that "infinity minus one" is sort of a weird notion but I don't understand what it means or why we can default it back to infinity, whereas in any other context when we change bounds we change them on both lower and upper.

Answer 1

The manipulations you did are valid for convergent series. Therefore your proof is incomplete because you didn't prove that the series is actually convergent.

Fortunately the convergence in this case is easy to show: The sequence of partial sums is strictly increasing (all terms of the series are positive), and each of the terms of that sequence is less than $1$. A strictly increasing sequence that is bound from above converges.

Here's a proof that shows that the partial sums are bounded by $1$ without knowledge of the result, by virtue of manipulations analogue to the manipulations you did in the infinite series:

Be $s_n = \sum_{k=1}^n \frac{9}{10^k}$ the $n$-th partial sum. Clearly, $s_1=\frac{9}{10}<1$. Furthermore, it is easy to check that $s_{n+1} = (s_n + 9)/10$, thus if $s_n<1$ then $s_{n+1} < (1+9)/10 = 1$. Therefore by induction, all $s_n<1$.

To see why establishing convergence is important, consider the following series: $$S = 1 + 1 + 1 + 1 + \ldots$$ Obviously, $$S = 0 + S = 0 + 1 + 1 + 1 + \ldots$$ But then, \begin{aligned} 0 &= S - S\\ &= (1-0) + (1-1) + (1-1) + (1-1) + \ldots\\ &= 1 + 0 + 0 + 0 + \ldots\\ &= 1 \end{aligned}

So why can you multiply the terms of the series by $10$ (or actually any number $c$) and get $10$ times the result?

Well, the usual definition of convergence of a series is as follows:

$\sum_{k=0}^{\infty} a_k = a$ if for every $\epsilon > 0$ you can find an $N_\epsilon$, so that for each $n>N_\epsilon$ you have $\left|\sum_{k=0}^n a_k - a\right| < \epsilon$.

So now to prove that $\sum_{k=0}^{\infty} ca_k = ca$, you have to find an $N'_\epsilon$, so that for each $n>N'_\epsilon$ you have $\left|\sum_{k=0}^n ca_k - ca\right| < \epsilon$. However note that the latter is now a finite product, and thus you can move the factor $c$ out of the sum. For $c=0$, the statement is obviously true, so we only have to consider $c\ne 0$. But then, we can reformulate the condition as $\left|\sum_{k=0}^n a_k - a\right| < \epsilon/\left|c\right|$, and since $\sum a_k$ converges to $a$, and thus we see that $N'_\epsilon = N_{\epsilon/\left|c\right|}$ works.

The case of the shift is even easier: You effectively have replaced the series $$a_0 + a_1 + a_2 + \ldots$$ by the series $$0 + a_0 + a_1 + a_2 + \ldots$$ Adding $0$ obviously doesn't change the finite sum, and then we see that $N'_\epsilon = N_\epsilon+1$ covers the exact same terms for the modified sequence.

You've also used a third operation which you didn't specifically mention: You subtracted the two series element-wise. The proof that this is allowed is similar to the previous ones, but slightly more complicated because you've now got two series. But again, you use the fact that you can do the manipulation for finite sums, and the fact that both series individually converge.

Answer 2

This is just one of the limit laws expressed in series form: if $c \in \mathbb R$ and $\displaystyle \sum_{k=1}^\infty a_k$ is a convergent series, then $\displaystyle \sum_{k=1}^\infty c a_k$ is also convergent and moreover $$\sum_{k=1}^\infty c a_k = c \sum_{k=1}^\infty a_k.$$

Answer 3

The formula for a finite geometric series is

$$\sum_{j=0}^k q^j=1+q+\cdots +q^k=\frac{1-q^{k+1}}{1-q}$$

In the case $|q|<1$ the term $q^{k+1}$ tends to $0$, if $k$ tends to $\infty.$ Therefore, the sum converges and has value $\frac{1}{1-q}$ in this case because the denominator is not $0$.

Your concrete example leads to $$\frac{9}{10}(1+\frac{1}{10}+\frac{1}{100}+\cdots)=\frac{9}{10}\frac{1}{1-\frac{1}{10}}=\frac{9}{10}\cdot \frac{1}{\frac{9}{10}}=\frac{9}{10}\cdot\frac{10}{9}=1$$

Answer 4

More simple, you have $$0.999\cdots = 0.999\cdots$$ then you multiply by ten on both sides

$$ 10 \times 0.999\cdots = 9.999\cdots=9+0.999\cdots$$ then substract $0.999\cdots$ from both sides

$$10 \times 0.9999\cdots-0.9999\cdots=(10-1)\times 0.999\cdots = 9$$ but $10-1=9$ so simplify $9$ from both sides and you get the equality $0.999\cdots=1$