4

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.

Xander Henderson
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AAron
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    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 $\frac{i(10^k-1)}{9}$. – lulu Jul 07 '15 at 01:37
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    It can be concisely expressed using the symbol $\infty$. – dalastboss Jul 07 '15 at 01:42
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    @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. – MJD Jul 07 '15 at 12:02
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    @MJD I believe that was intended as something of a joke, and a rather funny one at that. – Jared Smith Jul 07 '15 at 14:33

4 Answers4

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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 $\infty$

Conrado Costa
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7

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

corindo
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0

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: $$ \sum_{n=1}^{\infty}\sum_{m=1}^{n}(n \text{ mod } 10)*10^{m-1} $$

Travis
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0

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

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

Integrand
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Nissim Levy
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  • Can you provide the derivation of your summation formula please? – DDS Jan 15 '21 at 02:16
  • I will provide my derivation tonight or tomorrow. Is there a way to attach a text document to this comment? – Nissim Levy Jan 15 '21 at 19:54
  • 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. – DDS Jan 15 '21 at 21:39
  • 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:

     – Nissim Levy Jan 16 '21 at 21:48
  • 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

     – Nissim Levy Jan 16 '21 at 21:48
  • = Sum(n[(10^(n-k) - 1)/9]) - Sum(k[(10^(n-k) - 1)/9]) k = 0 to n-1 k = 0 to n-1

    n is a constant, so we can put it outside the summation, but K isn't. So we can write:

    n/9 * Sum(10^(n-k) - 1) - 1/9 * Sum(k10^(n-k) - k) k = 0 to n-1 k = 0 to n-1

    The sum formula for the first summation is easily found so I won't show it here. The second term is more tricky becasue it's comprised of k multiplied by 10^(n-k) and these two cannot be separated.

     – Nissim Levy Jan 16 '21 at 21:48
  • So how can we get the sum of k10^(n-k)|k = 0 to n-1?

    sum = 10^(n-1) + 210^(n-2) + 3 10^(n-3) + ... + k * 10^(n-k) +...+ (n-1)10 1/10Sum = 10^(n-2) + 210^(n-3) + 3 10^(n-4) + ... + k * 10^(n-k-1) +...+ (n-1)

    Sum - 1/10Sum = 10^(n-1) + 10^(n-2) + 10^(n-3) +...+ 10^(n-k) + 10 - n + 1 1/10(Sum - 1/10Sum) = 10^(n-2) + 10^(n-3) + 10^(n-4) +...+ 10^(n-k - 1) + 1 -n/10 + 1/10 1/10(Sum - 1/10Sum) = (10^(n-1) - 1)/9 -n/10 + 1/10

    and now we just need to solve for the sum and add it into the whole expression whose steps I skipped.

     – Nissim Levy Jan 16 '21 at 21:49
  • I should make it clear that the summation equation produced from this appraoch does not use modulus 10. In other words the tenth term is not 0000000000 The tenth term is 10* 1111111111 The kth term is k * k 1s – Nissim Levy Jan 16 '21 at 21:51
  • Thank you for posting your derivation; a suggestion---it would be easier for the reader if you would add it to your existing answer---which can be accomplished by clicking on the edit button and continuing from there. Thank you for all this effort. – DDS Jan 16 '21 at 22:05
  • I realize my contribution looks very messy. The full derivation is very long and it would be difficult to present it here in a legible way. – Nissim Levy Jan 16 '21 at 22:42