Sum of Sequence of Reciprocals of 4 n + 1 Alternating in Sign

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Theorem

\(\ds \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac 1 {4 n + 1}\) \(=\) \(\ds 1 - \frac 1 5 + \frac 1 9 - \frac 1 {13} + \frac 1 {17} - \cdots\)
\(\ds \) \(=\) \(\ds \dfrac {\pi \sqrt 2} 8 + \dfrac {\sqrt 2 \map \ln {1 + \sqrt 2} } 4\)


Proof

From Primitive of Power and the Fundamental Theorem of Calculus, we have:

$\ds \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac 1 {4 n + 1} = \sum_{n \mathop = 0}^\infty \paren {-1}^n \int_0^1 x^{4 n} \rd x$

We can rewrite:

\(\ds \sum_{n \mathop = 0}^\infty \paren {-1}^n \int_0^1 x^{4 n} \rd x\) \(=\) \(\ds \lim_{N \mathop \to \infty} \sum_{n \mathop = 0}^N \paren {-1}^n \int_0^1 x^{4 n} \rd x\) Definition of Infinite Series
\(\ds \) \(=\) \(\ds \lim_{N \mathop \to \infty} \int_0^1 \paren {\sum_{n \mathop = 0}^N \paren {-1}^n x^{4 n} } \rd x\) Linear Combination of Definite Integrals

We now use Lebesgue's Dominated Convergence Theorem to swap integral and summation.

Specifically, we will prove that:

$\ds \lim_{N \mathop \to \infty} \int_0^1 \paren {\sum_{n \mathop = 0}^N \paren {-1}^n x^{4 n} } \rd x = \int_0^1 \paren {\lim_{N \mathop \to \infty} \sum_{n \mathop = 0}^N \paren {-1}^n x^{4 n} } \rd x$

From Alternating Series Test: Lemma, we have:

$\ds \size {\sum_{n \mathop = 0}^N \paren {-1}^n x^{4 n} } \le 1$

for all $x \in \closedint 0 1$ and $N \in \N$.

All functions involved are continuous, so the conditions for Lebesgue's Dominated Convergence Theorem are satisfied, and we have:

$\ds \lim_{N \mathop \to \infty} \int_0^1 \paren {\sum_{n \mathop = 0}^N \paren {-1}^n x^{4 n} } \rd x = \int_0^1 \paren {\lim_{N \mathop \to \infty} \sum_{n \mathop = 0}^N \paren {-1}^n x^{4 n} } \rd x$

Then, by the definition of infinite series, we have:

$\ds \int_0^1 \paren {\lim_{N \mathop \to \infty} \sum_{n \mathop = 0}^N \paren {-1}^n x^{4 n} } \rd x = \int_0^1 \paren {\sum_{n \mathop = 0}^\infty \paren {-x^4}^n} \rd x$

Then we have:

\(\ds \int_0^1 \paren {\sum_{n \mathop = 0}^\infty \paren {-x^4}^n} \rd x\) \(=\) \(\ds \int_0^1 \frac 1 {1 + x^4} \rd x\) Sum of Infinite Geometric Sequence
\(\ds \) \(=\) \(\ds \intlimits {\frac 1 {2 \sqrt 2} \paren {\map \arctan {\frac 1 {\sqrt 2} \paren {x - \frac 1 x} } + \frac 1 2 \ln \size {\frac {x^2 + \sqrt 2 x + 1} {x^2 - \sqrt 2 x + 1} } } } 0 1\) Primitive of $\dfrac 1 {1 + x^4}$
\(\ds \) \(=\) \(\ds \frac 1 {2 \sqrt 2} \intlimits {\map \arctan {\frac 1 {\sqrt 2} \paren {x - \frac 1 x} } } 0 1 + \frac 1 {4 \sqrt 2} \intlimits {\ln \size {\frac {x^2 + \sqrt 2 x + 1} {x^2 - \sqrt 2 x + 1} } } 0 1\)

We have:

\(\ds \intlimits {\frac 1 {2 \sqrt 2} \map \arctan {\frac 1 {\sqrt 2} \paren {x - \frac 1 x} } } 0 1\) \(=\) \(\ds \frac 1 {2 \sqrt 2} \arctan 0 - \frac 1 {2 \sqrt 2} \lim_{x \mathop \to 0^+} \map \arctan {\frac 1 {\sqrt 2} \paren {\frac {x^2 - 1} x} }\)
\(\ds \) \(=\) \(\ds -\frac 1 {2 \sqrt 2} \lim_{x \mathop \to 0^+} \map \arctan {\frac 1 {\sqrt 2} \paren {\frac {x^2 - 1} x} }\) Arctangent of Zero is Zero
\(\ds \) \(=\) \(\ds -\frac 1 {2 \sqrt 2} \lim_{u \mathop \to -\infty} \map \arctan {\frac 1 {\sqrt 2} u}\) setting $u = x - \dfrac 1 x$
\(\ds \) \(=\) \(\ds \frac \pi {4 \sqrt 2}\) Limit to Infinity of Arctangent Function
\(\ds \) \(=\) \(\ds \frac {\pi \sqrt 2} 8\)

and:

\(\ds \frac 1 {4 \sqrt 2} \intlimits {\ln \size {\frac {x^2 + \sqrt 2 x + 1} {x^2 - \sqrt 2 x + 1} } } 0 1\) \(=\) \(\ds \frac 1 {4 \sqrt 2} \paren {\ln \size {\frac {2 + \sqrt 2} {2 - \sqrt 2} } - \ln 1}\)
\(\ds \) \(=\) \(\ds \frac 1 {4 \sqrt 2} \ln \size {\frac {2 + \sqrt 2} {2 - \sqrt 2} }\) Natural Logarithm of 1 is 0
\(\ds \) \(=\) \(\ds \frac 1 {4 \sqrt 2} \ln \size {\frac {\paren {2 + \sqrt 2}^2} {\paren {2 - \sqrt 2} \paren {2 + \sqrt 2} } }\)
\(\ds \) \(=\) \(\ds \frac {\sqrt 2} 8 \ln \size {\paren {\frac {2 + \sqrt 2} {\sqrt 2} }^2}\) Difference of Two Squares
\(\ds \) \(=\) \(\ds \frac {\sqrt 2 \map \ln {1 + \sqrt 2} } 4\) Logarithm of Power

Putting these together gives:

$\ds \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac 1 {4 n + 1} = \int_0^1 \frac 1 {1 + x^4} \rd x = \frac {\pi \sqrt 2} 8 + \frac {\sqrt 2 \map \ln {1 + \sqrt 2} } 4$

$\blacksquare$


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