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I would like to express

$$1+22+333+4444+\cdots$$

using $\Sigma$ notation, and have no clue where to start.

After $999999999$, comes 10 $0$s, then 11 $1$s.

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    $\begingroup$ Well, it appears useful to note that a string of, say, k ones can conveniently be expressed as $\frac{10^k-1}{9}$ . Similarly, a string of k i's (for any non-zero digit i) is just $\frac{i(10^k-1)}{9}$. $\endgroup$
    – lulu
    Commented Jul 7, 2015 at 1:37
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    $\begingroup$ It can be concisely expressed using the symbol $\infty$. $\endgroup$
    – dalastboss
    Commented Jul 7, 2015 at 1:42
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    $\begingroup$ @dalastboss That is not what the question is asking. An expression has a sense and a value. The expressions value is $\infty$, but the question is not what the value is but how to express the sense. If you don't distinguish between these two notions, you are in the uncomfortable position of saying (for example) that the statement $e^{i\pi} = -1$ is trivial, since of course the two sides have the same value. What makes this and similar statements interesting is that the two sides have different senses. This distinction goes back to Gottlob Frege. $\endgroup$
    – MJD
    Commented Jul 7, 2015 at 12:02
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    $\begingroup$ @MJD I believe that was intended as something of a joke, and a rather funny one at that. $\endgroup$ Commented Jul 7, 2015 at 14:33

4 Answers 4

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The terms of the sequence are

$$a_n =\left(n-10\cdot\left\lfloor\frac{n}{10}\right\rfloor\right)\cdot\frac{10^{n}-1}{9}$$

the sum of your series is $\infty$

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$$\sum_{n=1}^{\infty}\left(n-10\cdot\left\lfloor\frac{n}{10}\right\rfloor\right)\cdot\frac{10^{n}-1}{9}$$

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Considering $$ 4444 = 4*10^{3} + 4*10^{2} + 4*10^{1} + 4*10^{0} $$

I think a double sum and modulo is a lot more intuitive: $$ \sum_{n=1}^{\infty}\sum_{m=1}^{n}(n \text{ mod } 10)*10^{m-1} $$

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I've derived the formula that gives the sum of the series for $n$ terms. For example the series:

$$1 + 22 + 333 + 4444,$$ has four terms.

Let $n$ be the number of terms. Let $S_n$ be the sum of the $n$ terms. Then

$$S_n = \frac{1}{1458}\left((18n-2)10^{n+1} -81n^2 -81n + 20 \right)$$

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  • $\begingroup$ Can you provide the derivation of your summation formula please? $\endgroup$
    – DDS
    Commented Jan 15, 2021 at 2:16
  • $\begingroup$ I will provide my derivation tonight or tomorrow. Is there a way to attach a text document to this comment? $\endgroup$ Commented Jan 15, 2021 at 19:54
  • $\begingroup$ There is a limit to the number of characters that a comment will take; perhaps you might add it as an edit to your answer? Many thanks. $\endgroup$
    – DDS
    Commented Jan 15, 2021 at 21:39
  • $\begingroup$ I started by realizing the following: Sn = S{n-1} + n(1 + 10 + 100 +...+10^(n-1)) where S{n-1} is the sum up to the (n-1)th term S{n-1} = S{n-2} + (n-1)(1 + 10 + 100 + ... + 10^(n-2)) So Sn = S{n-2} + (n-1)(1 + 10 + 100 + ... + 10^(n-2)) + n(1 + 10 + 100 +...+10^(n-1)) Also, 1 + 10 + 100 +...+10^n = (10^(n+1) - 1)/9 So Sn = S{n-2} + (n-1)[(10^(n-1) - 1)/9] + n[(10^n - 1)/9] I kept substituting these S{n-k} equivalencies for S{n-k-1} until I got to S1 = 1 I ended up with the following series: $\endgroup$ Commented Jan 16, 2021 at 21:48
  • $\begingroup$ Sn = n[(10^n - 1)/9] + (n-1)[(10^(n-1) - 1)/9] + (n-2)[(10^(n-2) - 1)/9] +...+ (n-k)[(10^(n-k) - 1)/9] +...+ 1 This is expressed as Sum( (n-k)[(10^(n-k) - 1)/9] ) k = 0 to n-1 In the above summation n is a constant term and k is the variable over which the summation is performed. So Sn = Sum( (n)[(10^(n-k) - 1)/9] - (k)[(10^(n-k) - 1)/9] ) k = 0 to n-1 $\endgroup$ Commented Jan 16, 2021 at 21:48

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