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I was just curious for the proof/theorem for a very close decimal (so when you keep adding a decimal) to equal the next integer such as .999999999999....=1. Thank you.

isabeld99
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  • It's always important to Google to see if math.stackexchange has already answered your question. (This particular question is one of the most repeatedly asked math questions on the internet.) – littleO Mar 21 '21 at 01:09
  • The number $0.\bar{9}$ is by definition the limit of the sequence $.9, .99, .999, \ldots$, and clearly the limit of that sequence is $1$. – littleO Mar 21 '21 at 01:12

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The given number is a geometric series in disguise:

\begin{align*} 0.999\ldots = 9\sum_{n=1}^{\infty}10^{-n} = \frac{9}{10}\left(\frac{1}{1 - \frac{1}{10}}\right) = 1 \end{align*} which converges because its ratio is $0.1 < 1$.

user0102
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  • I'm not really satisfied with this answer as 1 is not really sum of $0.9+0.99+...$. Actually there is nothing like sum of infinite terms if we use the word "sum" as we do for finite sums. The word "Sum" in case of infinite series represents "limit" to which the series converges to. Therefore, $0.999... \to 1$ as per this answer but claiming that they are equal using convergence of geometric series, is just not correct. PS: I have not downvoted. – Koro Mar 21 '21 at 01:51
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    Whether you like infinite sums or not, they are well-defined, and $0,3333...$ or $0.9999,,,$ are only defined if you accept the definition of infinite sums. @Koro – Thomas Andrews Mar 21 '21 at 01:54
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    @Koro I think this is completely valid. Otherwise, how else do you define a recurring decimal? It really is just an infinite sum in disguise – Riemann'sPointyNose Mar 21 '21 at 01:55
  • Right @Riemann'sPointyNose and Thomas. – Koro Mar 21 '21 at 03:10