PMI (not neccesarily) Problem

To prove that 132+152+172++1(2n+1)2<14\frac{1}{3^2} + \frac{1}{5^2} + \frac{1}{7^2} + \cdots +\frac{1}{(2n+1)^2} < \frac{1}{4} for all n belongs to Nn\ belongs\ to\ N

This was given to be proved by induction in "Self-Learning Exercises - GSEB".

Note by Megh Parikh
5 years, 7 months ago

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Overkill: Here \sum indicates i=1\sum_{i=1}^\infty. Since 1i2=π26\sum \frac{1}{i^2} = \frac{\pi^2}{6}, we have 1(2i)2=π224\sum \frac{1}{(2i)^2} = \frac{\pi^2}{24}. Subtracting the two, we have 1(2i1)2=π28\sum \frac{1}{(2i-1)^2} = \frac{\pi^2}{8}, thus 1(2i+1)2=1(2i1)2112=π281<14\sum \frac{1}{(2i+1)^2} = \sum \frac{1}{(2i-1)^2} - \frac{1}{1^2} = \frac{\pi^2}{8} - 1 < \frac{1}{4}.

I'm pretty sure the method is to figure out some always positive f(n)f(n) such that f(n)1(2n+1)2>f(n+1)f(n) - \frac{1}{(2n+1)^2} > f(n+1), so we can induct it on i=1n1(2i+1)2<14f(n)\sum_{i=1}^n \frac{1}{(2i+1)^2} < \frac{1}{4} - f(n).

Ivan Koswara - 5 years, 7 months ago

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That is indeed an overkill. The problem is direct if you know that fact, and hence lets look for solutions which do not directly involve it.

Any suggestions for what f(n)f(n) could be?

Calvin Lin Staff - 5 years, 7 months ago

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By guesswork I obtained one f to be f(n)=n+1916(2n+1)2f(n)=\dfrac{n+\frac{19}{16}}{(2n+1)^2}

Megh Parikh - 5 years, 7 months ago

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Explaining my guesswork:

I assumed f to be rational function with denominator to be (2n+1)^2 and numerator to be linear ax+b

Then I got quadratic in n and made it into linear and got a=1.

Then solving for b in term of n, I got b>9/8

Also f(1)f(n+1)<1/4    b<5/4f(1) - f(n+1) < 1/4 \implies b<5/4

So I took b=9.5/8

Megh Parikh - 5 years, 7 months ago

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@Megh Parikh Nicely done.

Note that the second condition is superfluous, because we can choose a different starting point. For example, if we take b=2 b = 2 , then with g(n)=n+2(2n+1)2 g(n) = \frac{n+2}{(2n+1)^2} , we get that g(n)g(n+1)>1(2n+1)2 for n1. g(n) - g(n+1) > \frac{ 1}{ (2n+1) } ^2 \text{ for } n \geq 1 .

It remains to find an nn which would give us

132+152++1(2n+1)2<14g(n). \frac{1}{3^2 } + \frac{1}{5^2} + \ldots + \frac{1}{(2n+1)^2} < \frac{1}{4} - g(n) .

In this particular case, n=5 n = 5 works (using a calculator).

Calvin Lin Staff - 5 years, 7 months ago

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I didn't understand the subtraction of the summation. How did you get the denominator (2i1)2{(2i-1)}^2 ? .Could you explain better this part?

Carlos E. C. do Nascimento - 5 years, 7 months ago

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1i2=112+122+132+142+\sum \frac{1}{i^2} = \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \ldots, and 1(2i)2=122+142+162+\sum \frac{1}{(2i)^2} = \frac{1}{2^2} + \frac{1}{4^2} + \frac{1}{6^2} + \ldots, so subtracting the two gives 112+132+152+=1(2i1)2\frac{1}{1^2} + \frac{1}{3^2} + \frac{1}{5^2} + \ldots = \sum \frac{1}{(2i-1)^2}.

Ivan Koswara - 5 years, 7 months ago

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Thanks

Megh Parikh - 5 years, 7 months ago

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112+132+152+=(112+122+132+)(122+142+162+)=π26π246=π28\frac11^2+\frac13^2+\frac15^2+\dots=\left(\frac11^2+\frac12^2+\frac13^2+\dots\right)-\left(\frac12^2+\frac14^2+\frac16^2+\dots\right)=\frac{\pi^2}{6}-\frac{\pi^2}{4\cdot6}=\frac{\pi^2}{8}

Hence,

132+152+=π281<14\frac13^2+\frac15^2+\dots=\frac{\pi^2}{8}-1<\frac14

Since f(n)f(n) is increasing monotonically and the inequality holds asymptotically, the inequality holds for each nn.

Cody Johnson - 5 years, 7 months ago

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That is exactly my solution above, which is also an overkill (you have to prove 1i2=π26\sum \frac{1}{i^2} = \frac{\pi^2}{6}, for example).

Ivan Koswara - 5 years, 7 months ago

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