How to find sum of $n$ terms of a series with increasing difference?

Question

Find the sum up to $n$ terms of the series $0.4+0.44+0.444+\cdots$

I know how to find the sum up to n terms. I tried to getting the answer with the equation $a+(n-1)d$, but since the difference is increasing I am not sure.

So, here $a=0.4$ and $d=dn-d$

where $n$ is the difference between the adjacent terms.So anyway, I'm just guessing here, is $d=\frac{1}{10}$

Answer 1

One way to do this to break up each term into its individual digits and sum all of the ones with the same digit. In other words, you have

$$0.4 + 0.44 + 0.444 + \cdots = 0.4 + (0.4 + 0.04) + (0.4 + 0.04 + 0.004) + \cdots \tag{1}\label{eq1A}$$

There are $n$ terms of $0.4$, $n - 1$ terms of $0.04$, $n - 2$ terms of $0.004$, etc. Thus, the overall sum would be

$$\begin{equation}\begin{aligned} & \sum_{i=1}^{n}4(10^{-i})(n - i + 1) \\ & = \sum_{i=1}^{n}4(10^{-i})(n + 1) - \sum_{i=1}^{n}4(10^{-i})i \\ & = 4(n + 1)\sum_{i=1}^{n}10^{-i} - 4\sum_{i=1}^{n}i(10^{-i}) \\ & = 4(n + 1)(10^{-1})\left(\frac{1 - (10^{-1})^{n}}{1 - 10^{-1}}\right) - \\ & 4\left(\frac{10^{-1} - (1 + n)(10^{-1})(10^{-1})^{n}}{1 - 10^{-1}} + \frac{(10^{-1})(10^{-1})(1 - (10^{-1})^{n})}{(1 - 10^{-1})^2}\right) \\ & = 4(n + 1)\left(\frac{1 - (0.1)^{n}}{9}\right) - \\ & 4\left(\frac{0.1 - (1 + n)(0.1)^{n+1}}{0.9} + \frac{0.01(1 - 0.1^{n})}{0.81}\right) \end{aligned}\end{equation}\tag{2}\label{eq2A}$$

I'll leave it to you to further simplify the expression above. The first summation in the third line in \eqref{eq2A} is a geometric series, while the second summation is of a Arithmetico–geometric sequence, where I used the stated closed-form formula of

$$S_{n} = {\frac {ab-(a+nd)\,br^{n}}{1-r}}+{\frac {dbr\,(1-r^{n})}{(1-r)^{2}}}$$

with $a = d = 1$ and $b = r = 10^{-1}$.

Answer 2

Here is another way which is typically useful in similar problems, based on self similarity after multiplying/dividing by some number. But at first, notice that if we multiply $0.44\dots4$ (with $n$ fours) by $9/4$, we get $0.99\dots9$, which is just $1-1/10^n$. Now write \begin{align} x&=& &\underbrace{0.4+0.44+0.444+\dots}_{n-1 \text{ terms}}+0.\underbrace{44\dots4}_{n\text{ fours}}\\ 10x &=&4+&\underbrace{4.4+4.44+4.444+\dots}_{n-1 \text{ terms}} \end{align} By subtracting upper from lower, we get $9x=$$4+4+ \dots+4-0.44\dots4=$$4n-\frac{4}{9}(1-1/10^n)$, and expression for $x$ follows.