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About $1\over 1$, $1\over 2$, $1\over 3$, and $1\over 4$, can $1\over 4$ also be written as $1\over 5^1$+$1\over 5^2$+$1\over 5^3$+$1\over 5^4$+...=$1\over 5$+$1\over 25$+$1\over 125$+$1\over 625$+...? I think so because it's equal to

$0.2$

$0.04$

$0.008$

$0.016$

$0.0032$

$0.00064$

$0.000128$

$\dots$

<p>=$0.2499968$</p>

which is approximately equal to $0.25$=$1\over 4$. It just keeps going on forever. I think this can also happen for the other fractions at the beginning of this question ($1\over 3$=$1\over 4^1$+$1\over 4^2$+$1\over 4^3$+...=$1\over 4$+$1\over 16$+$1\over 64$+...) and other fractions that don't have a denominator of zero. What do you think?

Adola
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ReliableMathBoy
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3 Answers3

2

Yes. The sum of a decreasing geometric series is $$\frac{\text{first term}}{1-\text{common ratio}}$$

So in your examples $$\sum_{n=1}^\infty \frac{1}{(x+1)^n}=\frac{\frac{1}{x+1}}{1-\frac{1}{x+1}}=\frac{\frac{1}{x+1}}{\frac{(x+1)-(1)}{x+1}}=\frac{\frac{1}{x+1}}{\frac{x}{x+1}}=\frac{1}{x}\\\sum_{n=1}^\infty \frac{1}{5^n}=\frac{\frac{1}{5}}{1-\frac{1}{5}}=\frac{\frac{1}{5}}{\frac{(5)-(1)}{5}}=\frac{\frac{1}{5}}{\frac{4}{5}}=\frac{1}{4}$$

turkeyhundt
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Let's denote the sum $\frac{1}{5^1}+\frac{1}{5^2}+\frac{1}{5^3}+...$ as $x$. Now note that $\frac{1}{5}x=\frac{1}{5^2}+\frac{1}{5^3}+\frac{1}{5^4}+...$, so $\frac{1}{5}x+\frac{1}{5^1}=\frac{1}{5^1}+\frac{1}{5^2}+\frac{1}{5^3}+\frac{1}{5^4}+...=x$, so we have $\frac{1}{5}x+\frac{1}{5}=x$. Now, solving for $x$, we get $\frac{1}{5}=\frac{4}{5}x$, or $\frac{1}{4}=x$.

Wojowu
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Just to prove turkeyhundt's formula, we have:

$$\begin{align} \sum_{i=1}^{\infty}a^{i}&= a + \sum_{i=2}^{\infty}a^{i}\\ &= a + \sum_{n=1}^{\infty}a^{n+1}\\ &= a + a\sum_{n=1}^{\infty}a^{n}\\ \sum_{i=1}^{\infty}a^{i} - a \sum_{i=1}^{\infty}a^{i} &= a\\ \sum_{i=1}^{\infty}a^{i} &= \frac{a}{1-a} \end{align}$$

Which is the result. Note that in the second step, we said i = n+1, so i = 2 meant that n = 1, and then in the fourth step, we just relabeled n as i.

also, note that we can then look for a geometric sum for any fraction $1/k$, and we have:

$$\begin{align} \frac{1}{k} &= \frac{a}{1-a}\\ k &= \frac{1-a}{a}\\ &= \frac{1}{a} - 1\\ k+1 &= \frac{1}{a}\\ a &= \frac{1}{k+1}\\ \end{align}$$

which matches your guess.

Zo the Relativist
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    Note that for the full proof, you need to show that the series converge first. This does not work if $\lvert a\rvert\geq 1$. – tomasz May 04 '17 at 10:35
  • @tomasz: I think it should be pretty clear that $\sum a^n$ does not converge for $|a| \geq 1$ – Zo the Relativist May 04 '17 at 13:28
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    Of course. But one needs to note that the series does converge, or the calculation becomes meaningless as of line 4. – tomasz May 08 '17 at 00:40