Ok...
Let #P_n# be the proposition that #sum_{k=1}^{2n} (1/{k(k+1)} + (-1)^k) geq 2/3#
First check that it holds for #P_1# where #n=1#
#P_1: sum_{k=1}^{2} (1/{k(k+1)} + (-1)^k) geq 2/3#
#(1/{1(1+1)} + (-1)^1)+(1/{2(2+1)} + (-1)^2) geq 2/3#
If you're doing mathematical induction I'm going to assume you don't need help with basic algebra so I'll leave out the steps for simplifying the equation.
#-1/2 + 7/6 geq 2/3#
#2/3 geq 2/3#
As we can see, #P_1# is true.
The next step is to assume #P_k# is true, but as k is already used in the expression I'm going to write it as #P_m#. We assume #P_m# is true:
#P_m: sum_{k=1}^{2m} (1/{k(k+1)} + (-1)^k) geq 2/3#
Now we need to prove that #P_{m+1}# is true provided that #P_m# is true.
#P_{m+1}: sum_{k=1}^{2(m+1)} (1/{k(k+1)} + (-1)^k) geq 2/3#
#P_{m+1}: sum_{k=1}^{2m+2} (1/{k(k+1)} + (-1)^k) geq 2/3#
This sum can be split into two parts:
#P_{m+1}: sum_{k=1}^{2m+2} (1/{k(k+1)}) + sum_{k=1}^{2m+2}((-1)^k) geq 2/3#
Let's take a look at the part with #(-1)^k#
#sum_{k=1}^{2m+2}((-1)^k) = (-1)^1 + (-1)^2+(-1)^3+...+(-1)^{2m}+(-1)^{2m+1}+(-1)^{2m+2} #
#sum_{k=1}^{2m+2}((-1)^k) = -1 + 1 - 1+...+((-1)^m)^2+(-1)^{2m+1}+((-1)^{m+1})^2 =#
Writing #((-1)^{m+1})^2# makes it clear that we are raising #(-1)^{m+1}# to the power of 2 and will give us 1 regardless of the sign of #(-1)^{m+1}#
We will get alternating #-1# and #+1#, and since it terminates with #+1# (as there are an even amount of terms) the sum will be 0.
#sum_{k=1}^{2m+2}((-1)^k) = -1 + 1 - 1+...+1-1+1=0#
So...
#P_{m+1}: sum_{k=1}^{2m+2} (1/{k(k+1)} + (-1)^k)=sum_{k=1}^{2m+2} (1/{k(k+1)}) geq 2/3#
This sum differs by 2 terms from #P_m# so we can rewrite this as:
#sum_{k=1}^{2m+2} (1/{k(k+1)}) =sum_{k=1}^{2m} (1/{k(k+1)}) +(1/{(2m+1)((2m+1)+1)})+(1/{(2m+2)((2m+2)+1)})#
Earlier we assumed #P_m# to be true:
#sum_{k=1}^{2m} (1/{k(k+1)} + (-1)^k) geq 2/3#
We can split this up as well:
#sum_{k=1}^{2m} (1/{k(k+1)} )+ sum_{k=1}^{2m}( (-1)^k) geq 2/3#
And since the number of terms is even #sum_{k=1}^{2m}((-1)^k) = -1 + 1 - 1+...+1-1+1=0#
Thus
#P_m: sum_{k=1}^{2m}(1/{k(k+1)} ) geq 2/3#
Since this is assumed to be true we can substitute this into the expression as #2/3 + a# #{a in ZZ⁺}# where a is some constant since it is greater than or equal to #2/3#. Including #+a# may or may not be necessary but it shows that it could be #2/3# or #2/3# plus some number that is greater.
#2/3 +a +(1/{(2m+1)((2m+1)+1)})+(1/{(2m+2)((2m+2)+1)}) geq 2/3#
Since #m in ZZ^{+}# it follows that the other two terms are positive.
#m > 0 => (1/{(2m+1)((2m+1)+1)})+(1/{(2m+2)((2m+2)+1)})>0#
Let these two terms be equal to #b# where #b>0#
#2/3 +a +b geq 2/3# where #a,b > 0#
Since #a geq 0# and #b > 0#, #2/3 +a +b# must also be greater than #2/3#.
So.........
Since #P_n# holds for #P_1# and #P_{m+1}# whenever #P_m# is true, #P_n# holds for all #n in ZZ^{+}#
