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Expert: Robi Bhattacharjee - 9/23/2008


Question
Prove by mathematical induction that

SQRT(2)/2 + SQRT(3)/4 + SQRT(4)/6 +...+ SQRT(n+1)/2n > SQRT(n)/2 for all intergers n >= 1.

Answer
Hello Michael,

For this problem, it is obvious that the case n = 1 works. Now all we have to do is that assuming it works for the j, we must prove it works for n = j + 1.

The j + 1 th term in this sequence is sqrt(j +2)/2(j + 1). This is what we are adding to the left side of the inequality. Now the right side just transforms into sqrt(j+1)/2. So this is the same as adding sqrt(j+1)/2 - sqrt(j)/2 to the left side of the equation.

Now there is a rule in inequalities that if a > b and c > d then
a + c > b + d.

So we know that the rest of the series up to the jth term is bigger than the sqrt(j)/2 by our induction assumption. So if we can show that
sqrt(j +2)/2(j + 1) is > sqrt(j+1)/2 - sqrt(j)/2 then we are finished with our proof.

So let us just write sqrt(j + 2)/2(j + 1) > sqrt(j+1)/2 - sqrt(j)/2

first let us multiply both sides by 2. That will just get rid of the 2's in the bottom.

now if we multiply both sides by sqrt(j+1) + sqrt(j), we get

(sqrt(j+1) + sqrt(j)))sqrt(j + 2)/(j + 1) > 1

so...........

(sqrt(j+1)sqrt(j + 2) + sqrt(j)sqrt(j+2)) / (j + 1) > 1

this is true since sqrt(j+1)sqrt(j + 2) is > (j+1) since (j+1) =

sqrt(j+1)*sqrt(j+1).

So it is proven.

Please ask if anything wasn't clear.  


Robi  

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Robi Bhattacharjee

Expertise

I can answer a variety of questions on mathematics. Questions on trigonometry, calculus(preferably single variable), algebra, geometry, and number theory will be answered. I cannot answer questions on abstract branches of mathematics such as group theory. I also cannot answer questions on statistics. In number theory, I can answer questions on congruences, prime numbers, units, functions, and the riemann-zeta function.

Experience

I have studied advanced math my entire life. I started calculus in sixth grade. I have attended numerous math competitions and I am attending math organizations such as the San-Diego math circle. Also, this year I have been invited to the USAMO which is a prestigious math competition (Every year the USAMO invites 500 students from across the USA to participate in this competition. The top 6 go to represent the USA in the International Math Olympiad).

Organizations
I am in the San Diego Math Circle

Education/Credentials
I am entering high school and have received a perfect score and the STAR test 5 times in a row. I also have gotten recognitions in the AMC 10, AIME, Math Counts, and ARML. Additionally, I have won the San Diego Math Olimpiad twice in a row.

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