Proof by induction?

Relevant wiki: Geometric Progression Sum

$\begin{aligned} S & = \frac a{a+1} + \left(\frac a{a+1}\right)^2 + \left(\frac a{a+1}\right)^3 + \cdots + \left(\frac a{a+1}\right)^n \\ & = \sum_{k=1}^n \left(\frac a{a+1}\right)^k & \small \color{#3D99F6} \text{Let }r = \frac a{a+1} \\ & = \sum_{k=1}^n r^k = r \left(\frac {r^n-1}{r-1}\right) & \small \color{#3D99F6} \text{A sum of geometric progression} \\ & = \frac {15}{16} \times \frac {1-\left(\frac {15}{16}\right)^{67}}{1-\frac {15}{16}} & \small \color{#3D99F6} \text{For }r = \frac {15}{16} \text{ and }n=67 \\ & \approx \boxed{14.801} \end{aligned}$

Let's start by finding the general formular for the series: $\frac{a}{a+1}+(\frac{a}{a+1})^{2}+(\frac{a}{a+1})^{3}+(\frac{a}{a+1})^{4}+...+(\frac{a}{a+1})^{n}$

The way we can start is by setting $a = 1.$ When we do this, we find a more similar series: $\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+...+(\frac{1}{2})^{n}$ .

This series is equal to $\frac{2^{n}-1}{2^{n}} \Leftrightarrow \frac{2^{n}-1^{n}}{2^{n}}$

We may think here that $\frac{a}{a+1}+(\frac{a}{a+1})^{2}+(\frac{a}{a+1})^{3}+(\frac{a}{a+1})^{4}+...+(\frac{a}{a+1})^{n}$ ) is equal to $\frac{(a+1)^{n}-a^{n}}{(a+1)^{n}}$

But before we begin to prove it, let's also test it for $a = 2$

If $n = 1$ then the series will just be $\frac{2}{3}$ , but by using $\frac{(a+1)^{n}-a^{n}}{(a+1)^{n}}$ we will get the result of $\frac{1}{3}$ . This is two times smaller.

If $n = 2$ then the series will be $\frac{2}{3} + (\frac{2}{3})^{2}$ . This is equal to $\frac{10}{9}$ . However by using $\frac{(a+1)^{n}-a^{n}}{(a+1)^{n}}$ we get $\frac{5}{9}$ which again is two times smaller.

Let's try setting $a = 3$

If $n = 1$ then the series will just be $\frac{3}{4}$ , but by using $\frac{(a+1)^{n}-a^{n}}{(a+1)^{n}}$ we will get the result of $\frac{1}{4}$ . This is 3 times smaller.

We can see the general pattern here, that $\frac{(a+1)^{n}-a^{n}}{(a+1)^{n}}$ should be changed to $\frac{a((a+1)^{n}-a^{n})}{(a+1)^{n}}$ .

So it seems likely that, $\frac{a}{a+1}+(\frac{a}{a+1})^{2}+(\frac{a}{a+1})^{3}+(\frac{a}{a+1})^{4}+...+(\frac{a}{a+1})^{n}$ = $\frac{a((a+1)^{n}-a^{n})}{(a+1)^{n}}$

We can prove this by induction, however I'll leave that an exercise to the reader.

When you've proven it, all we need to do is plug in $a = 15$ and $n = 67$ .

Thus we get that, $\frac{15}{16}+(\frac{15}{16})^{2}+(\frac{15}{16})^{3}+(\frac{15}{16})^{4}+...+(\frac{15}{16})^{67}$ = $\frac{15(16^{67}-15^{67})}{16^{67}}$

And this is the step, where you'll probably need a calculator. By typing the equation into a calculator you get the result that:

$\frac{15(16^{67}-15^{67})}{16^{67}} = 14.801$

This means, that: $\frac{15}{16}+(\frac{15}{16})^{2}+(\frac{15}{16})^{3}+(\frac{15}{16})^{4}+...+(\frac{15}{16})^{67} = 14.801$