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.