Please i need help , using the Mathematical induction ?

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1 Answer
Apr 21, 2018

See method below.

Explanation:

Ok...

Let P_nPn be the proposition that sum_{k=1}^{2n} (1/{k(k+1)} + (-1)^k) geq 2/32nk=1(1k(k+1)+(1)k)23

First check that it holds for P_1P1 where n=1n=1

P_1: sum_{k=1}^{2} (1/{k(k+1)} + (-1)^k) geq 2/3P1:2k=1(1k(k+1)+(1)k)23

(1/{1(1+1)} + (-1)^1)+(1/{2(2+1)} + (-1)^2) geq 2/3(11(1+1)+(1)1)+(12(2+1)+(1)2)23

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 2/312+7623

2/3 geq 2/32323

As we can see, P_1P1 is true.

The next step is to assume P_kPk is true, but as k is already used in the expression I'm going to write it as P_mPm. We assume P_mPm is true:

P_m: sum_{k=1}^{2m} (1/{k(k+1)} + (-1)^k) geq 2/3Pm:2mk=1(1k(k+1)+(1)k)23

Now we need to prove that P_{m+1}Pm+1 is true provided that P_mPm is true.

P_{m+1}: sum_{k=1}^{2(m+1)} (1/{k(k+1)} + (-1)^k) geq 2/3Pm+1:2(m+1)k=1(1k(k+1)+(1)k)23

P_{m+1}: sum_{k=1}^{2m+2} (1/{k(k+1)} + (-1)^k) geq 2/3Pm+1:2m+2k=1(1k(k+1)+(1)k)23

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 2/3Pm+1:2m+2k=1(1k(k+1))+2m+2k=1((1)k)23

Let's take a look at the part with (-1)^k(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^{+}