Prove using induction: $\sum_{i=1}^{n}\frac{(-2)^i}{i}\leq 2^{n-2}$ $\forall n \in \mathbb{N}$

I already checked it holds for $n=1$.

So let's suppose $\sum_{i=1}^{n}\frac{(-2)^i}{i}\leq 2^{n-2}$ holds for $n \in \mathbb{N}$, then we want to show that $\sum_{i=1}^{n+1}\frac{(-2)^i}{i}\leq 2^{n-1}$.

Using induction step i got to:

$2^{n-2} + \frac{(-2)^{n+1}}{n+1} \leq 2^{n-1}$

And this is where I'm stuck, I've tried to approach this problem from many different angles and none seem to work.

I would appreciate hint, thanks.

Solution 1:

Well as you mentioned correctly, we have

$$\sum_{i=1}^{n+1}\frac{(-2)^i}{i} = \sum_{i=1}^{n}\frac{(-2)^i}{i} + \frac{(-2)^{n+1}}{n+1}\leq 2^{n-2} + \frac{(-2)^{n+1}}{n+1}$$

And we have to prove

$$2^{n-2} + \frac{(-2)^{n+1}}{n+1} \le 2^{n-1}$$

But this is equivalent to (divide both sides by $2^{n-2}$)

$$1 + \frac{(-1)^{n+1}2^3}{n+1} \le 2 \Rightarrow \frac{(-1)^{n+1}2^3}{n+1} \le 1 \Rightarrow \frac{(-1)^{n+1}}{n+1} \le \frac{1}{2^3} $$

Which holds true for $n\ge 6$. Your first step of induction that you have to check is $n=1,2,3,4,5$, but they hold true.Here is a list

$$\begin{align} & n=1 : -2 \le \frac{1}{2}\\ & n=2 : 0 \le 2 \\ & n=3 : \frac{-8}{3} \le 4 \\ & n=4 : \frac{4}{3} \le 8 \\ & n=5 : \frac{-4}{5} \le 16 \end{align}$$