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2

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

We have:

$1=(10-1)/9,$

$11=(10^2-1)/9$

.

.

.

$11...11=(10^n-1)/9$

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

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

Also the general form of terms is:

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

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)dn ;[0, 1]$

but it does not work. Can someone says what is wrong i.e why the integral does

not give $S$ as I mentioned first?

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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. – peterh 9 hours ago
3  
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. – leonbloy 9 hours ago
2  
Agreed, I'm not entirely sure how you went from the summation notation to the integral notation, changing the bounds when you did. – Cort Ammon 8 hours ago
up vote 16 down vote

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

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