Sum of infinite series.

Question

If $$S=\frac{2}{10}+\frac{6}{10^2}+\frac{12}{10^3}+\frac{20}{10^4}+\cdots,$$ find the value of $S$.

1. $24/90$
2. $242/900$
3. $245/900$
4. $200/729$

What I understand:

Had the numerators not been present, it could have been solved by infinite geometric series formula. Also, I can't find any relationship between the numerators. Please help me with this question.

Answer 1

Very elegant solution from Math Lover indeed, let me provide you a solution that's a bit more hands on:

$$S = \frac{2}{10} + \frac{6}{10^2} + \frac{12}{10^3} +\frac{20}{10^4} + \ ...$$ Multiply both sides by the factor $\frac{1}{10}$:

$$\frac{S}{10} = \frac{2}{10^2} + \frac{6}{10^3} + \frac{12}{10^4} +\frac{20}{10^5} + \ ...$$

Evaluate $S - \frac{S}{10}$:

$$S - \frac{S}{10} = \frac{2}{10} + \Big(\frac{6}{10^2} - \frac{2}{10^2}\Big) + \Big(\frac{12}{10^3} - \frac{6}{10^3}\Big) +\Big(\frac{20}{10^4} - \frac{12}{10^4}\Big) + \ ...$$ $$S - \frac{S}{10} = \frac{2}{10} + \frac{4}{10^2} + \frac{6}{10^3} +\frac{8}{10^4} + \ ...$$

And let's multiply this result by $\frac{1}{10}$:

$$\frac{S}{10} - \frac{S}{10^2} = \frac{2}{10^2} + \frac{4}{10^3} + \frac{6}{10^4} +\frac{8}{10^5} + \ ...$$

And let's subtract $\frac{S}{10} - \frac{S}{10^2}$ from $S - \frac{S}{10}$ using the same logic:

$$\Bigg(S - \frac{S}{10}\Bigg) - \Bigg(\frac{S}{10} - \frac{S}{10^2} \Bigg) = \frac{2}{10} + \Big(\frac{4}{10^2} - \frac{2}{10^2}\Big) + \Big(\frac{6}{10^3} - \frac{4}{10^3}\Big) + \Big(\frac{8}{10^4} - \frac{6}{10^4}\Big) + \ ...$$ $$\Bigg(S - \frac{S}{10}\Bigg) - \Bigg(\frac{S}{10} - \frac{S}{10^2} \Bigg) = \frac{2}{10} + \frac{2}{10^2} + \frac{2}{10^3} +\frac{2}{10^4} + \ ...$$

And now let's multiply by $\frac{1}{10}$ to get a new sum, then we subtract the new sum from the old sum.. again. This time however we can see that all of the terms cancel each other out except for the first fraction $\frac{2}{10}$: $$\Bigg(\Bigg(S - \frac{S}{10}\Bigg) - \Bigg(\frac{S}{10} - \frac{S}{10^2} \Bigg)\Bigg) - \Bigg(\frac{\Bigg(S - \frac{S}{10}\Bigg) - \Bigg(\frac{S}{10} - \frac{S}{10^2} \Bigg)}{10}\Bigg)\Bigg) = \frac{2}{10}$$

Now simplify that mess and we get:

$$\frac{729}{10^3}S = \frac{2}{10}$$ $$729S = 200$$ $$S = \frac{200}{729}$$

So the correct answer is (4): $\frac{200}{729}$

Answer 2

When $|x|<1$, $$\frac{1}{1-x} = 1+x+x^2+x^3+x^4+x^5+\cdots$$ Differentiate both sides twice to get $$\frac{2}{(1-x)^3} = 2+6x+12x^2 + 20 x^3+ \cdots$$ Consequently, $$\frac{2x}{(1-x)^3} = 2x+6x^2+12x^3 + 20 x^4+ \cdots$$ Put $x=1/10$ to find the required sum.

Answer 3

Here's how to get a more general result with just algebra (no calculus).

Let $s_m(x) =\sum_{n=0}^{\infty} p_m(n)x^n $, where $|x| < 1$ (so all the series converge) and $p_m(n) =\prod_{k=1}^m (n+k-1) =n(n+1)...(n+m-1) $.

I will show that, for $m \ge 1$, $s_m(x) =\dfrac{m!x}{(1-x)^{m+1}} $.

In particular, $\sum_{n=1}^{\infty} \dfrac{n(n+1)}{10^n} =s_2(1/10) =\dfrac{2(1/10)}{(9/10)^3} =\dfrac{200}{729} $.

We have

$\begin{array}\\ p_m(n+1)-p_m(n) &=\prod_{k=1}^m (n+k)-\prod_{k=1}^m (n+k-1)\\ &=\prod_{k=1}^m (n+k)-\prod_{k=0}^{m-1} (n+k)\\ &=(n+m)\prod_{k=1}^{m-1} (n+k)-n\prod_{k=1}^{m-1} (n+k)\\ &= (n+m-n)\prod_{k=1}^{m-1}(n+k)\\ &=mp_{m-1}(n+1)\\ \end{array} $

Therefore, for $m \ge 1$,

$\begin{array}\\ (1-x)s_m(x) &=\sum_{n=0}^{\infty} p_m(n)x^n-x\sum_{n=0}^{\infty} p_m(n)x^n\\ &=\sum_{n=0}^{\infty} p_m(n)x^n-\sum_{n=0}^{\infty} p_m(n)x^{n+1}\\ &=\sum_{n=0}^{\infty} p_m(n)x^n-\sum_{n=1}^{\infty} p_m(n-1)x^{n}\\ &=p_m(0)+\sum_{n=1}^{\infty} (p_m(n)-p_m(n-1))x^n\\ &=\sum_{n=1}^{\infty} mp_{m-1}(n)x^n \qquad\text{since }p_m(0) = 0\\ &=m\sum_{n=1}^{\infty} p_{m-1}(n)x^n\\ &=m(-p_{m-1}(0)+\sum_{n=0}^{\infty} p_{m-1}(n)x^n)\\ &=m(-p_{m-1}(0)+s_{m-1}(x))\\ &=ms_{m-1}(x)-mp_{m-1}(0)\\ &=ms_{m-1}(x)-1 \qquad\text{for } m=1\\ &=ms_{m-1}(x) \qquad\text{for }m \ge 2\\ \end{array} $

We have $s_0(x) =\dfrac1{1-x} $.

Therefore $(1-x)s_1(x) =s_0(x)-1 =\dfrac{x}{1-x} $ so $s_1(x) =\dfrac{x}{(1-x)^2} $.

For $m \ge 2$, $s_m(x) =\dfrac{ms_{m-1}(x)}{1-x} $

From this, $s_2(x) =\dfrac{2s_{1}(x)}{1-x} =\dfrac{2x}{(1-x)^3} $.

By an easy induction, $s_m(x) =\dfrac{m!x}{(1-x)^{m+1}} $.