Solving an unusual equation

Question

I need to find a real number $n$ such that $n > 1$ and:

$$ \sum_{k=1}^\infty \frac{2^k}{n^k} = \frac{n-1}{n} $$

Ideally, I'd find the minimum such $n$ (if more than one exists), but really, any such $n$ will do.

I cannot for the life of me figure out how to solve this equation for $n$! Are there any strategies I might be missing?

Answer 1

$$\sum_{k=1}^{\infty}\frac{2^k}{n^k}=\left(\frac{2}{n}\right)+\left(\frac{2}{n}\right)^2+\left(\frac{2}{n}\right)^3+\cdots+\left(\frac{2}{n}\right)^\infty$$

Common Ratio is $\frac{2}{n}$ So, This will converge when $|n|>2$.

$$\sum_{k=1}^{\infty}\frac{2^k}{n^k}=\frac{\frac{2}{n}}{1-\frac{2}{n}}$$

$$\sum_{k=1}^{\infty}\frac{2^k}{n^k}=\frac{\frac{2}{n}}{1-\frac{2}{n}}=\frac{2}{n-2}$$

It's all easy then $$\frac{2}{n-2}=\frac{n-1}{n}$$ gives quadratic in n $$n^2-5n+2=0$$ since $n>2$ we get $$n=\frac{5+\sqrt{17}}{2}$$

Answer 2

We can begin by considering the series:

$$\frac{1}{1-x} = \sum_{k=0}^\infty{ x^k } \tag{1}$$

$$\frac{1}{1-x} - 1 = \sum_{k=1}^\infty{ x^k } \tag{2}$$

Then let $x=\frac{2}{n}$:

$$\frac{1}{1-\left(\frac{2}{n}\right)} - 1 = \sum_{k=1}^\infty{ \left(\frac{2}{n}\right)^k } \tag{3}$$

Now we have the desired sum, and we want this equal to $\frac{n-1}{n}$:

$$\frac{1}{1-\left(\frac{2}{n}\right)} - 1 = \frac{n-1}{n} \tag{4}$$

You can probably guess how to proceed from here. Wolfram Alpha gives these answers:

Solving for $n$, we get:

$$n=\begin{cases} \frac{5+\sqrt{17}}{2} \\ \frac{5-\sqrt{17}}{2} \end{cases} \tag{5}$$

Of course, the greater value is $\frac{5+\sqrt{17}}{2}$.