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

$b.1+\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+.....$

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marked as duplicate by mfl, Hans Lundmark, Arjang, LutzL, Martin Sleziak Dec 1 '16 at 12:38

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  • $\begingroup$ Write each series in summation form and use either the ratio or root tests. $\endgroup$ – mattos Dec 1 '16 at 6:46
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    $\begingroup$ Do you want the value of the sum or to know where they are converging as you state in your question? They are two different things. $\endgroup$ – mattos Dec 1 '16 at 6:49
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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.

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Hint

$$1+\frac{1}{9}+\frac{1}{25}+\frac{1}{49}+\cdots=1+\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+\frac{1}{25}+\frac{1}{36}+\frac{1}{49}+\cdots-(\frac{1}{4}+\frac{1}{16}+\frac{1}{36}+\cdots)$$ Now $$\frac{1}{4}+\frac{1}{16}+\frac{1}{36}+\cdots=\frac{1}{4}(1+\frac{1}{4}+\frac{1}{9}+\cdots)$$

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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 $$\sum_{n=1}^{\infty}\frac{1}{n^2} - \sum_{n=1}^{\infty}\frac{1}{(2n)^2} = \frac{3}{4}\sum_{n=1}^{\infty}\frac{1}{n^2} = \frac{\pi^2}{8}.$$

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  • $\begingroup$ How do you get $\pi/6$? Both sums are greater than $1$. $\endgroup$ – Mike Pierce Dec 1 '16 at 6:52
  • $\begingroup$ @MikePierce it is well known. $\endgroup$ – Jacob Wakem Dec 1 '16 at 6:53
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    $\begingroup$ 1. 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 $\zeta(2)\neq\pi/6$, $\zeta(2) = \pi^2/6$. $\endgroup$ – Stahl Dec 1 '16 at 6:55
  • $\begingroup$ Modified the answer. Please check your calculations before posting $\endgroup$ – vidyarthi Dec 1 '16 at 7:18
  • $\begingroup$ How the value is to be known? What is the proof? $\endgroup$ – Arbin Dec 1 '16 at 7:48

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