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If $$a_{n}=\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+....+\frac{1}{(2n+1)^2}$$where $n\in N$.

Then prove that

$a_{n}<\frac{1}{4}$

Maverick
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4 Answers4

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$$\sum_{n\geq 1}\frac{1}{(2n+1)^2}<\sum_{n\geq 1}\frac{1}{2n(2n+2)}\stackrel{\text{telescopic!}}{=}\frac{1}{4}. $$ Since the LHS equals $\frac{\pi^2}{8}-1$ by Fourier series, this turns out to be a proof of $\pi^2<10$.

Jack D'Aurizio
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Here's a sketch of a direct proof that only requires calculus. Note that for each $k$, we have $$\frac{1}{(2k + 1)^2} \leq \int_{k-1}^k \frac{1}{(2x + 1)^2}\,dx.$$

This implies that each $a_n$ is less than $$\frac{1}{9} + \frac{1}{25} + \frac{1}{49} + \int_{3}^\infty \frac{1}{(2x + 1)^2}.$$

Calculating this integral directly gives the upper bound of $$\frac{1}{9} + \frac{1}{25} + \frac{1}{49} + \frac{1}{14} \approx .242 < \frac{1}{4}.$$

This method is a good technique for computing an upper bound on a sum (or a corresponding lower bound by altering the integral).

Marcus M
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The $n$th term is less than $\frac{1}{4n(n+1)}$. This replaces your series with a standard series that sums to $\frac{1}{4}$.

Math Lover
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  • Welcome to math SE. If you know latex then write the equations as you do in latex. – Math Lover Nov 13 '17 at 18:30
  • Thanks Math Lover. I really must learn how to write equations in something other than Word. –  Nov 13 '17 at 18:39
  • I've edited the mathematical expressions for you. I hope you'll figure out how to write math expressions/equations in mathjax or latex. – Math Lover Nov 13 '17 at 18:42
  • Thanks again - I'll refrain from answering until I have learnt. –  Nov 13 '17 at 18:44
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    It shouldn't discourage you posting your answers. Writing mathematical expressions in latex/mathjax is very easy. You can easily learn that. Here is a tutorial https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – Math Lover Nov 13 '17 at 18:45
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adding and subtracting $1 + \frac{1}{2^2} + \frac{1}{4^2} + \ldots + \frac{1}{(2n)^2}$, and using Basel problem we get $\lim_{n\to \infty}a_n = \sum_{k=1}^{2n+1} \frac{1}{k^2} - \frac{1}{4}\left(\sum_{k=1}^{n} \frac{1}{k^2}\right) - 1 = \frac{\pi^2}{6} - \frac{1}{4}\frac{\pi^2}{6} - 1 = \frac{\pi^2}{8} - 1$

sku
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