Proving the Inductive Formula for Series of Fractions using Mathematical Induction

In this article, we will prove the given statement for all natural numbers n using mathematical induction:

1 - 1/2 1/3 - 1/4 ... 1/(2n-1) - 1/2n 1/n1 1/n2 ... 1/(2n)

We will break down each step in the proof to ensure a clear understanding of the process.

Base Case

We begin by checking the base case when n 1:

1 - 1/2 1/1 1/2

The left side simplifies to:

1 - 1/2 1/2

Both sides are equal, so the base case holds true.

Inductive Step

Next, we assume that the statement holds for some natural number k:

1 - 1/2 1/3 - 1/4 ... 1/(2k-1) - 1/2k 1/k1 1/k2 ... 1/(2k)

This is our induction hypothesis. Now we need to prove that the statement also holds for k 1:

1 - 1/2 1/3 - 1/4 ... 1/(2k-1) - 1/2k 1/(2k 1) - 1/(2k 2) 1/(k 1)1 1/(k 1)2 ... 1/(2k 2)

The left side can be rewritten as:

(1 - 1/2 1/3 - 1/4 ... 1/(2k-1) - 1/2k) × (1/(2k 1) - 1/(2k 2))

Using the induction hypothesis, we substitute:

(1/k1 1/k2 ... 1/(2k)) × (1/(2k 1) - 1/(2k 2))

Now, we need to simplify the right side of the equation:

1/(k 1)1 1/(k 1)2 ... 1/(2k 2)

Notice that:

The product (1/k1 1/k2 ... 1/(2k) 1/(2k 1) - 1/(2k 2)) gives us all terms from 1/(k 1)1 to 1/(2k 2).

The term -1/(2k 2) is exactly the last term needed to complete the sequence to 1/(k 1)1 1/(k 1)2 ... 1/(2k 2).

Therefore, the inductive step holds.

Conclusion

By the principle of mathematical induction, the statement is proven for all natural numbers n:

1 - 1/2 1/3 - 1/4 ... 1/(2n-1) - 1/2n 1/n1 1/n2 ... 1/(2n)

This concludes our proof of the inductive formula for series of fractions using mathematical induction.