Method of mathematical induction?

Question

Mathematics

Help please, for the second day I have been suffering with these inequalities, if you wish, n still cannot be equalized to one

Answer 1

Perhaps the condition was rewritten incorrectly, or an error in the textbook. The first formula is wrong.

1/3 != 1/2
1/3+1/15 = 2/5 != 2/3
1/3+1/15 + 1/35 = 3/7 != 3/4

The correct formula
is sum = n/(2n+1)
Now it can already be proved inductively. The base is obvious. Transition:
Previous sum 1/3+...1/(4n^2-1) = n/(2n+1)
Add the following term:
n/(2n+1) + 1/(4(n+1)^2 - 1) =carefully decompose the second denominator according to the difference of squares formula, reduce to a common denominator and factorize the numerator of the = (n+1)/(2n+3)FTD.
The second inequality can be proved. The base is obvious (1-a1)(1-a2) = 1-a1-a2-a1*a2 > 1-a1-a2because a1*a2- is positive. Next, take the product for n, replace the first n-1 parentheses by the induction hypothesis, open the parentheses, and then you can omit all terms like ai*aj, they are all positive.