2 Geometric Progressions

Relevant wiki: Geometric Progression Sum

$1+\dfrac{1}{r}+\dfrac{1}{r^2}+\dfrac{1}{r^3}+\dfrac{1}{r^4} + \ldots = 1-\dfrac{1}{s}+\dfrac{1}{s^2}-\dfrac{1}{s^3}+\dfrac{1}{s^4} - \ldots$

Notice that the LHS is the sum of a geometric progression to infinity, where $a=1$ and $r=\dfrac{1}{r}$

Similarly, notice that the RHS is also the sum of another geometric progression to infinity, where $a=1$ and $r=-\dfrac{1}{s}$

Therefore, we can replace these two sides of the equation with the formula for sums of GP to infinity:

$\dfrac{1}{1-\frac{1}{r}} = \dfrac{1}{1-\left(-\frac{1}{s}\right)}\\ \dfrac{1}{\frac{r-1}{r}} = \dfrac{1}{\frac{s+1}{s}}\\ \dfrac{r}{r-1} = \dfrac{s}{s+1}\\ r(s+1)=s(r-1)\\ rs+r=rs-s\\ \boxed{r+s=0}$

Let $s = -x$ . The alternating sum becomes the sum of all inverse powers of $x$ , with no alternating signs; compare it with the series for $r$ , and you can match every power of $x$ with every power of $r$ ; thus, $r = x$ , which means $r = -s$ , or $r + s = 0$ .

The common ratio for the L.H.S of the equation is $\frac{1}{r}÷1=\frac{1}{r}$ , while that of the R.H.S of the equation is $\frac{-1}{s}÷1=\frac{-1}{s}$ .

So, sum of the geometric infinite series in L.H.S of the equation= $\frac{1}{1-\frac{1}{r}}=\frac{r}{r-1}$ .

And the sum of the series in the R.H.S= $\frac{1}{(1+\frac{1}{s})}=\frac{s}{s+1}$ .

So, $L.H.S=R.H.S$

$\Rightarrow \frac{r}{r-1}=\frac{s}{s+1}$

$\Rightarrow rs+r=rs-s$

$\Rightarrow r+s=0$

The other answers are great, but I'll leave my two cents here as well.

$1 + \dfrac{1}{r} + \dfrac{1}{r^{2}} + \dfrac{1}{r^{3}} + \ldots = 1 - \dfrac{1}{s} + \dfrac{1}{s^{2}} - \dfrac{1}{s^{3}} + \ldots$

We can try to write the progression using sigma notation:
$\displaystyle\sum_{k = 1}^{\infty} r^{-k} = \sum_{k = 1}^{\infty} -s^{-k}$

Then we use the formula for the sum of infinite geometric series:
$\dfrac{1}{1 - r^{-k}} = \dfrac{1}{1 + s^{-k}}$

Simplify a bit:
$1 - r^{-k} = 1 + s^{-k}$
$- r^{-k} = s^{-k}$

Transform it into logs:
$\log_{-r} s^{-k} = -k$

Use some log properties to simplify it even further:
$-k \cdot \log_{-r} s = -k$
$\log_{-r} s = 1$
$-r = s$

And now we arrive at the answer:
$r + s = 0$