To prove that \[\frac{1}{3^2} + \frac{1}{5^2} + \frac{1}{7^2} + \cdots +\frac{1}{(2n+1)^2} < \frac{1}{4} \] for all \(n\ belongs\ to\ N\)

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

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## Comments

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TopNewestOverkill: Here \(\sum\) indicates \(\sum_{i=1}^\infty\). Since \(\sum \frac{1}{i^2} = \frac{\pi^2}{6}\), we have \(\sum \frac{1}{(2i)^2} = \frac{\pi^2}{24}\). Subtracting the two, we have \(\sum \frac{1}{(2i-1)^2} = \frac{\pi^2}{8}\), thus \(\sum \frac{1}{(2i+1)^2} = \sum \frac{1}{(2i-1)^2} - \frac{1}{1^2} = \frac{\pi^2}{8} - 1 < \frac{1}{4}\).

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

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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 \(f(n) \) could be?

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

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

So I took b=9.5/8

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

It remains to find an \(n\) which would give us

\[ \frac{1}{3^2 } + \frac{1}{5^2} + \ldots + \frac{1}{(2n+1)^2} < \frac{1}{4} - g(n) .\]

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

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Thanks

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

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\(\sum \frac{1}{i^2} = \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \ldots\), and \(\sum \frac{1}{(2i)^2} = \frac{1}{2^2} + \frac{1}{4^2} + \frac{1}{6^2} + \ldots\), so subtracting the two gives \(\frac{1}{1^2} + \frac{1}{3^2} + \frac{1}{5^2} + \ldots = \sum \frac{1}{(2i-1)^2}\).

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\[\frac11^2+\frac13^2+\frac15^2+\dots=\left(\frac11^2+\frac12^2+\frac13^2+\dots\right)-\left(\frac12^2+\frac14^2+\frac16^2+\dots\right)=\frac{\pi^2}{6}-\frac{\pi^2}{4\cdot6}=\frac{\pi^2}{8}\]

Hence,

\[\frac13^2+\frac15^2+\dots=\frac{\pi^2}{8}-1<\frac14\]

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

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

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