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".

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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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 \(\sum_{i=1}^n \frac{1}{(2i+1)^2} < \frac{1}{4} - f(n)\). – Ivan Koswara · 3 years, 2 months ago

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Any suggestions for what \(f(n) \) could be? – Calvin Lin Staff · 3 years, 2 months ago

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– Megh Parikh · 3 years, 2 months ago

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

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 – Megh Parikh · 3 years, 2 months ago

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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). – Calvin Lin Staff · 3 years, 2 months ago

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– Megh Parikh · 3 years, 2 months ago

ThanksLog in to reply

– Carlos E. C. do Nascimento · 3 years, 2 months ago

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

– Ivan Koswara · 3 years, 2 months ago

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

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

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 \(n\). – Cody Johnson · 3 years, 2 months ago

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– Ivan Koswara · 3 years, 2 months ago

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).Log in to reply