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I know that the series b. converges as $\sum \frac{1}{n^p}$ converges for $p>1$, So a. also converges. I want to know the sum.

a.$1+\frac{1}{9}+\frac{1}{25}+\frac{1}{49}+.....$

<p>$b.1+\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+.....$</p>
Arbin
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3 Answers3

2

If you know $\displaystyle \sum_1^\infty \frac1{n^2} = \frac{\pi^2}6$, then this is easy, as the even series is just $\displaystyle \sum_1^\infty \frac1{4n^2} = \frac{\pi^2}{24}$.

I leave the odd series for you to separate out.

Macavity
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The Riemann zeta function is defined as $$\zeta(s) = \sum_{n=1}^{\infty}\frac{1}{n^s}.$$ The value of $\zeta(2)$ is known to be $\frac{\pi^2}{6}$. Thus $$\sum_{n=0}^{\infty}\frac{1}{n^2} = \frac{\pi^2}{6}.$$ The series in a. in your post can be written as $$\sum_{n=1}^{\infty}\frac{1}{n^2} - \sum_{n=1}^{\infty}\frac{1}{(2n)^2} = \frac{3}{4}\sum_{n=1}^{\infty}\frac{1}{n^2} = \frac{\pi^2}{8}.$$

vidyarthi
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Ethan Alwaise
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  • How do you get $\pi/6$? Both sums are greater than $1$. – Mike Pierce Dec 01 '16 at 06:52
  • @MikePierce it is well known. – Jacob Wakem Dec 01 '16 at 06:53
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  • Your sums should start at $n = 1$. 2. How are you deriving your expression for the series in part a? The series in part a appears to be $\sum_{n\textrm{ odd}}n^{-2}$, which is strictly less than $\zeta(2)$ (by comparison, all terms in the series of a appear in b, but not vice versa, and all are positive), but $1 + \pi/12 > \pi/6$. 3. @alephnull $\zeta(2)\neq\pi/6$, $\zeta(2) = \pi^2/6$.
    • – Stahl Dec 01 '16 at 06:55
  • Modified the answer. Please check your calculations before posting – vidyarthi Dec 01 '16 at 07:18
  • How the value is to be known? What is the proof? – Arbin Dec 01 '16 at 07:48