Sum of Reciprocals of Even Powers of Integers Alternating in Sign

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Theorem

Let $n \in \Z_{> 0}$ be a (strictly) positive integer.

\(\ds \sum_{j \mathop = 1}^\infty \paren {-1}^{j + 1} \frac 1 {j^{2 n} }\) \(=\) \(\ds \dfrac 1 {1^{2 n} } - \dfrac 1 {2^{2 n} } + \dfrac 1 {3^{2 n} } - \dfrac 1 {4^{2 n} } + \cdots\)
\(\ds \) \(=\) \(\ds \paren {-1}^{n + 1} \dfrac {B_{2 n} \paren {2^{2 n - 1} - 1} \pi^{2 n} } {\paren {2 n}!}\)

where $B_{2 n}$ is the $2 n$th Bernoulli number.


Corollary

Let $n \in \Z_{> 0}$ be a (strictly) positive integer.

\(\ds B_{2 n}\) \(=\) \(\ds \paren {-1}^{n + 1} \dfrac {\paren {2 n}!} {\paren {2^{2 n - 1} - 1} \pi^{2 n} } \sum_{j \mathop = 1}^\infty \paren {-1}^{j + 1} \frac 1 {j^{2 n} }\)
\(\ds \) \(=\) \(\ds \paren {-1}^{n + 1} \dfrac {\paren {2 n}!} {\paren {2^{2 n - 1} - 1} \pi^{2 n} } \paren {1 - \dfrac 1 {2^{2 n} } + \dfrac 1 {3^{2 n} } - \dfrac 1 {4^{2 n} } + \cdots}\)


Proof

\(\ds \sum_{j \mathop = 1}^\infty \paren {-1}^{j + 1} \frac 1 {j^{2 n} }\) \(=\) \(\ds \sum_{j \mathop = 1}^\infty \frac 1 {\paren {2 j - 1}^{2 n} } - \sum_{j \mathop = 1}^\infty \frac 1 {\paren {2 j}^{2 n} }\) separating odd positive terms from even negative terms
\(\ds \) \(=\) \(\ds \sum_{j \mathop = 1}^\infty \frac 1 {\paren {2 j - 1}^{2 n} } - \frac 1 {2^{2 n} } \sum_{j \mathop = 1}^\infty \frac 1 {j^{2 n} }\)
\(\ds \) \(=\) \(\ds \paren {-1}^{n + 1} \dfrac {B_{2 n} \paren {2^{2 n} - 1} \pi^{2 n} } {2 \paren {2 n}!} - \frac 1 {2^{2 n} } \sum_{j \mathop = 1}^\infty \frac 1 {j^{2 n} }\) Sum of Reciprocals of Even Powers of Odd Integers
\(\ds \) \(=\) \(\ds \paren {-1}^{n + 1} \dfrac {B_{2 n} \paren {2^{2 n} - 1} \pi^{2 n} } {2 \paren {2 n}!} - \frac 1 {2^{2 n} } \paren {-1}^{n + 1} \dfrac {B_{2 n} 2^{2 n - 1} \pi^{2 n} } {\paren {2 n}!}\) Riemann Zeta Function at Even Integers
\(\ds \) \(=\) \(\ds \paren {-1}^{n + 1} \dfrac {2^{2 n - 1} B_{2 n} \paren {2^{2 n} - 1} \pi^{2 n} - B_{2 n} 2^{2 n - 1} \pi^{2 n} } {2^{2 n} \paren {2 n}!}\)
\(\ds \) \(=\) \(\ds \paren {-1}^{n + 1} \dfrac {B_{2 n} 2^{2 n - 1} \paren {2^{2 n} - 2} \pi^{2 n} } {2^{2 n} \paren {2 n}!}\)
\(\ds \) \(=\) \(\ds \paren {-1}^{n + 1} \dfrac {B_{2 n} \paren {2^{2 n - 1} - 1} \pi^{2 n} } {\paren {2 n}!}\)

$\blacksquare$


Also see


Sources

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