Sum to $n$ terms of the given series

Question

Find the sum to $n$ terms of the given series:

$$0.3+0.33+0.333+0.3333+\cdots$$

My Attempt: Let $$S=0.3+0.33+0.333+0.3333+\cdots \text{ to $n$ terms}$$ $$=\frac {3}{10}+\frac {33}{100}+\frac {333}{1000} + \frac {3333}{10000}+\cdots$$ $$=\frac {3}{10} \left[1+\frac {11}{10}+\frac {111}{100}+\frac {1111}{1000}+\cdots \text{ to $n$ terms}\right]$$

How do I continue from here?

Answer 1

Multiply the sum by $3$ and add $0.111\cdots1$ ($n$ decimals). You get $n$, by distributing the ones.

But $0.111\cdots1=\dfrac19-\dfrac{10^{-n}}9$ (the subtrahend cancels the tail).

Hence

$$S=\frac{n-0.111\cdots1}3=\frac{9n-1+10^{-n}}{27}.$$

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Without words:

$$\begin{matrix}3\times&0.3&+0.33&+0.333&+0.3333\\ +&0.1&+0.01&+0.001&+0.0001\\ \hline\\ =&1&+1&+1&+1\end{matrix}$$

and

$$\begin{align}&0.1111111111111111\cdots\\ -&0.0000111111111111\cdots\\\hline =&0.1111 \end{align}$$

Answer 2

Write your sum to $n$ terms as $$S_n=0.3+(0,3+0.03)+(0.3+0.03+0.003)+\ldots +(0.3+0.03+0.003+\ldots 0.000000000003)$$ Now note that you have $n$ terms $0.3$, $n-1$ terms $0.03$, $n-2$ terms $0.003$ and so on, so $$S_n=0.3n+0.03(n-1)+0.003(n-2)+\ldots+3\cdot 10^{-n}(1)\\ S_n=\sum_{i=1}^n3\cdot 10^{-i}(n+1-i)$$ and we have lots of questions summing the arithmo-geometric series.

Answer 3

$$\dfrac {3}{10} \left(1+\dfrac {11}{10}+\dfrac {111}{100}+\dfrac {1111}{1000}+\ldots\right) \\ =\dfrac {3}{10} \left(\dfrac{1}{10^0}+\dfrac {10 +1}{10^1}+\dfrac {10^2 +10 +1}{10^2}+\dfrac {10^3+10^2+10+1}{10^3}+\ldots\right) $$ Each fraction has a (finite) geometric series in the denominator (there is a formula for those) and a power of ten in the denominator. $$=\frac3{10}\sum_{k=0}^{n-1} \frac1{10^k} \sum_{j=0}^k 10^j .$$

Answer 4

Here's different flavor that's more elementary and explicit (maybe).

At the $k$-th term among summing $k=1$ to $k = n$, we have the multiplicative factor you got: \begin{align*} \frac{ 111\cdots11 }{ 100\cdots00} &= 1 + \frac1{10} + \frac1{100} + \cdots \frac1{10^{k-1}} \tag*{, totally $k$ terms}\\ &= \frac{ 1 - (1/10)^k}{1 - 1/10} \\ &= \frac{10}9 \left( 1 - \frac1{10^k}\right) \end{align*} Thus the original summation of $n$ terms is \begin{align*} S = \frac3{10}\frac{10}9 \sum_{k = 1}^n \left( 1 - \frac1{10^k}\right) &= \frac13 \cdot \left(n - \frac{ \frac1{10} - \frac1{10^{n+1}} }{1 - \frac1{10}} \right) \\ &=\frac13 \cdot \left(n - \frac19 \left(1 - \frac1{10^n} \right) \right) \\ &= \frac13 \cdot \left(n - \frac19 \frac{ \,\,\overbrace{ 999\cdots 99 }^{n~\text{digits} }}{ \underbrace{ 100\cdots 000 }_{n+1~\text{digits} } } \right) \\ &= \frac13 \left( n - \frac{ \,\,\overbrace{ 111\cdots 11 }^{n~\text{digits} }}{ \underbrace{ 100\cdots 000 }_{n+1~\text{digits} } }\right) = \frac13 \cdot (n - 0.111\cdots 11) \end{align*} There are $n$ digits of $1$ following the decimal point. Or if you prefer,$$S = \frac13 \bigl(n - 1 + 0.888\cdots 89 \bigr)$$where there are $n-1$ digits of $8$ followed by a single $9$.

Answer 5

You need to calculate $$A=1+\dfrac {99\div 9}{10}+\dfrac {999\div9}{100}+\dfrac {9999\div 9}{1000}+...+\dfrac{(10^{n}-1)\div9}{10^{n-1}}$$

$$\begin{equation}\begin{aligned} A&=\frac {3}{10} \left(1+\frac {11}{10}+\frac {111}{100}+\frac {1111}{1000}+\cdots +\frac{(10^{n}-1)\div9}{10^{n-1}}\right) \\ &=1+\dfrac {99\div 9}{10}+\dfrac {999\div9}{100}+\dfrac {9999\div 9}{1000}+\cdots+\dfrac{(10^{n}-1)\div9}{10^{n-1}} \\ &=1+\dfrac{99\times10^{n-2}\div9}{10^{n-1}}+\dfrac{999\times10^{n-3}\div9}{10^{n-1}}+\cdots+\dfrac{(10^{n}-1)\div9}{10^{n-1}} \\ &=1+\dfrac{(10^2-1)\times10^{n-2}+(10^3-1)\times10^{n-3}+(10^4-1)\times10^{n-4}+\cdots+(10^n-1)}{9\times 10^{n-1}} \\ &=1+\dfrac{10^n-10^{n-2}+10^n-10^{n-3}+10^n-10^{n-4}+\cdots+10^n-1}{9\times 10^{n-1}}\text{ ($n-1$ terms)}\\ &=1+\dfrac{(n-1)10^n-(10^0+10^1+10^2+\cdots+10^{n-3}+10^{n-2})}{9\times 10^{n-1}} \end{aligned}\end{equation}$$

Continue to calculate

$$\begin{equation}\begin{aligned} B&=10^0+10^1+10^2+\cdots+10^{n-3}+10^{n-2} \\ 10B&=10^1+10^2+10^3+\cdots+10^{n-2}+10^{n-1}\\ \Rightarrow9B&=10B-B=10^{n-1}-1\\ B&=\dfrac{10^{n-1}-1}{9} \end{aligned}\end{equation}$$

We can conclude that

$$\begin{equation}\begin{aligned} A&=1+\dfrac{(n-1)10^n-\dfrac{10^{n-1}-1}{9}}{9\times 10^{n-1}} \end{aligned}\end{equation}$$

Answer 6

Let $$S_n=0.3+0.33+0.333+0.3333+\cdots \text{ to $n$ terms}$$ $$=\underbrace{\frac {3}{10}+\frac {33}{100}+\frac {333}{1000} + \frac {3333}{10000}+\cdots}_n$$ $$=3\left(\frac {1}{10}+\frac {11}{100}+\frac {111}{1000} + \frac {1111}{10000}+\cdots\right)$$ $$=3\left(\frac {1}{10^1}+\frac {11}{10^2}+\frac {111}{10^3} + \frac {1111}{10^4}+\cdots\right)$$ $$=3\left(\frac {\sum_{k=0}^010^k}{10^1}+\frac {\sum_{k=0}^110^k}{10^2}+\frac {\sum_{k=0}^210^k}{10^3} + \frac {\sum_{k=0}^310^k}{10^4}+\cdots+\frac {\sum_{k=0}^{(n-1)}10^k}{10^n}\right)$$

And $$\frac{1}{10^{n}}\left(\sum_{k=0}^{n-1}10^k\right)= \frac{1}{10^{n}}\left(\frac{10^{n}-1}{9}\right) = \frac{1}{9}\left(1 - \frac{1}{10^n}\right) = \frac{1}{9}\left(1 - 10^{-n}\right)$$ Therefore \begin{align} S_n &= 3\left(\sum_{k=1}^n\left[\frac{1}{9}\left(1 - 10^{-k}\right)\right]\right)\\ &= \frac{3}{9}\sum_{k=1}^n\left(1 - 10^{-k}\right)\\ &= \frac{1}{3} \left( \sum_{k=1}^n 1 - \sum_{k=1}^n10^{-k}\right)\\ &= \frac{1}{3} \left( n- \sum_{k=1}^n10^{-k}\right)\\ &= \frac{1}{3} \left( n- \frac{1}{9}\left(1 - 10^{-n}\right)\right) \end{align}

$$S_n = \frac{1}{27}\left(10^{-n}+ 9n -1\right) $$

Answer 7

The recursion $10S_n=S_{n-1}+3n$ with $S_0=0$ implies

$$S_n=a\left(\left(1\over10\right)^n-1\right)+bn$$

for some $a$ and $b$. From $S_1={3\over10}$ and $S_2={63\over100}$ we have

$$\begin{align} {3\over10}&=a\left({1\over10}-1\right)+b\\ {63\over100}&=a\left({1\over100}-1\right)+2b \end{align}$$

or

$$\begin{align} 9a-10b&=-3\\ 99a-200b&=-63 \end{align}$$

which solves to $b=1/3$ and $a=1/27$, so

$$S_n={1\over27}\left(\left(1\over10\right)^n-1\right)+{1\over3}n$$