Sum of series : $1+11+111+...$

Sum of series $1+11+111+\cdots+11\cdots11$ ($n$ digits)

We have:

$1=\frac {10-1}9,$

$11=\frac {10^2-1}9$

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$11...11= \frac {10^n-1}9$ (number with $n$ digits)

and summing them we find the sum ($S$) as:

$S=(10^{n+1}-9n-10)/81$

Also the general form of terms is:

$s(n)=(10^{n+1}-10^n-9)/81$

Now consider the function:

$f(x)=10^{x+1}−10^x-9$

Since $\Delta x= 1$, due to definition of integral we can write:

$S=(1/81)\sum (10^{x+1}−10^x-9), [1, ∞]$

$=(1/81)∫(10^{x+1}-10^x-9) dx ;[0, 1]$

but it does not work. Can someone say what went wrong, i.e, Why doesn't the integral give $S$ as I mentioned first?

I realized now that this is more a sequence rather than a series. A sequence is a set of numbers which are resulted from a general term where as a series is a set of functional elements; the derivative of elements of a sequence is zero and its integration is pointless. So using the integration of general term of a sequence to find its sum is just not needed.

• I would like to continue my answer, but I think there is a syntax error in your integral. Note, a little bit of latex knowledge would be useful to learn. Sep 29 '16 at 19:02
• An integral is not a sum. Sometimes one can express (the asympotic of) a sum as a Riemann sum, and hence reduce it to an integral. You cannot do that here. Sep 29 '16 at 19:15
• Agreed, I'm not entirely sure how you went from the summation notation to the integral notation, changing the bounds when you did. Sep 29 '16 at 19:46
• This is arithmetico-geometric series (progression, sequence). See Wikipedia or AoPS. You can find a very similar question here. Sep 30 '16 at 5:25
• Why is this question tagged 'geometric topology'? Oct 1 '16 at 14:40

\begin{align*} S &= \sum_{i=1}^n (10^i-1)/9 \\[6pt] &= \frac{1}{9} \left(\sum_{i=1}^n 10^i - \sum_{i=1}^n 1 \right) \\[6pt] &= \frac{1}{9} \left(\frac{10}{9}(10^n -1) - n\right) \\[6pt] &= \frac{10}{81} (10^n -1) - \frac{n}{9} \end{align*}