Before even starting with the induction proof, it looks like the expression has a useless term: we have
sum_{k=1}^{2n}(\frac{1}{k(k+1)}+(-1)^k) = sum_{k=1}^{2n}\frac{1}{k(k+1)}+sum_{k=1}^{2n}(-1)^k
Let's focus on the second sum. We are summing an even number of times (from 1 to 2n) the powers of -1, which sum to zero. For, example, we have
n=1: (-1)^1+(-1)^2 = -1+1=0
n=2: (-1)^1+(-1)^2+(-1)^3+(-1)^4= -1+1-1+1=0
and so on. So, the expression is the same if we ignore the second sum. So, let's define the proposition P(n) as
sum_{k=1}^{2n}\frac{1}{k(k+1)}\ge 2/3
and let's prove P(n) for all n \ge 1
For n=1, we have to prove that
sum_{k=1}^{2}\frac{1}{k(k+1)}\ge 2/3
We can explicitly write the first two terms:
\frac{1}{1(1+1)}+\frac{1}{2(2+1)} = 1/2+1/6 = 4/6=2/3 \ge 2/3
So, the base case P(1) holds.
Now, if we assume that P(n) is true for some n \in \mathbb{N}, we must show that P(n+1) holds as well.
We have that P(n+1) is the proposition
sum_{k=1}^{2(n+1)}\frac{1}{k(k+1)} = sum_{k=1}^{2n+2}\frac{1}{k(k+1)}\ge2/3
and we already know that
sum_{k=1}^{2n}\frac{1}{k(k+1)}\ge 2/3
This means that P(n+1) only has two terms more than P(n), and all terms are positive, since they are \frac{1}{k(k+1)}.
So, the sum up to 2n is already greater than 2/3, and we are adding two more (positive) terms, so we have
sum_{k=1}^{2(n+1)}\frac{1}{k(k+1)} \ge sum_{k=1}^{2n}\frac{1}{k(k+1)}\ge 2/3
Which proves the statement.