Proof by induction: For all $n \geq 1$; $1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \cdots +(-1)^{n+1} \frac{1}{n} \leq 1$ Proof by induction: For all $n \geq 1$; $1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \cdots +(-1)^{n+1} \frac{1}{n} \leq 1$
This is what I have so far:
Base case: for $n = 1$
$(-1)^{1+1} \cdot \frac{1}{1} \leq 1$
$(-1)^2 \leq 1$
$1 \leq 1$  
This holds true.
Inductive step: for $n = k$
Assume $1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \cdots + (-1)^{k+1} \frac{1}{k} \leq 1$ is true.
We want to show for $K=1$
$1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \cdots + (-1)^{k+1} \frac{1}{k} - (-1)^{k+2} \cdot \frac{1}{K+1} \leq 1$
$(-1)^{k+1} \frac{1}{k} - (-1)^{k+2} \cdot \frac{1}{K+1} \leq 1$
By doing some algebra I get this:
$-\dfrac{(-1)^k + (-1)^k k + (-1)^K k)}{k(1 + k)} \leq 1$
I am lost here, not sure if this says anything. Any help would be appreciated. Thanks in advance.

Hint: rewrite the sum as
$$
1-\left(\frac12-\frac13\right)-\left(\frac14-\frac15\right)-\left(\frac16-\frac17\right)-\dots
$$
