Answer on Question #85698 – Math – Real Analysis
Question
Obtain the value of x x x
∑ n = 1 ∞ 1 ⋅ 3 ⋅ 5 ⋅ … ⋅ ( 4 n − 3 ) 2 ⋅ 4 ⋅ 6 ⋅ … ⋅ ( 4 n − 2 ) ⋅ x 2 n 4 n \sum_{n=1}^{\infty} \frac{1 \cdot 3 \cdot 5 \cdot \ldots \cdot (4n-3)}{2 \cdot 4 \cdot 6 \cdot \ldots \cdot (4n-2)} \cdot \frac{x^{2n}}{4n} n = 1 ∑ ∞ 2 ⋅ 4 ⋅ 6 ⋅ … ⋅ ( 4 n − 2 ) 1 ⋅ 3 ⋅ 5 ⋅ … ⋅ ( 4 n − 3 ) ⋅ 4 n x 2 n
is convergent.
Solution
Use the Ratio Test
lim n → ∞ ∣ a n + 1 a n ∣ = lim n → ∞ ∣ 1 ⋅ 3 ⋅ 5 ⋅ … ⋅ ( 4 n − 3 ) ( 4 ( n + 1 ) − 3 ) 2 ⋅ 4 ⋅ 6 ⋅ … ⋅ ( 4 n − 2 ) ( 4 ( n + 1 ) − 2 ) ⋅ x 2 ( n + 1 ) 4 ( n + 1 ) ∣ = lim n → ∞ ∣ n ( 4 n + 1 ) x 2 ( n + 1 ) ( 4 n + 2 ) ∣ ∣ a n + 1 a n ∣ = ∣ n ( 4 n + 1 ) x 2 ( n + 1 ) ( 4 n + 2 ) ∣ → x 2 as n → ∞ \begin{array}{l}
\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{1 \cdot 3 \cdot 5 \cdot \ldots \cdot (4n-3)(4(n+1)-3)}{2 \cdot 4 \cdot 6 \cdot \ldots \cdot (4n-2)(4(n+1)-2)} \cdot \frac{x^2(n+1)}{4(n+1)} \right| \\
= \lim_{n \to \infty} \left| \frac{n(4n+1)x^2}{(n+1)(4n+2)} \right| \\
\left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{n(4n+1)x^2}{(n+1)(4n+2)} \right| \to x^2 \text{ as } n \to \infty
\end{array} lim n → ∞ ∣ ∣ a n a n + 1 ∣ ∣ = lim n → ∞ ∣ ∣ 2 ⋅ 4 ⋅ 6 ⋅ … ⋅ ( 4 n − 2 ) ( 4 ( n + 1 ) − 2 ) 1 ⋅ 3 ⋅ 5 ⋅ … ⋅ ( 4 n − 3 ) ( 4 ( n + 1 ) − 3 ) ⋅ 4 ( n + 1 ) x 2 ( n + 1 ) ∣ ∣ = lim n → ∞ ∣ ∣ ( n + 1 ) ( 4 n + 2 ) n ( 4 n + 1 ) x 2 ∣ ∣ ∣ ∣ a n a n + 1 ∣ ∣ = ∣ ∣ ( n + 1 ) ( 4 n + 2 ) n ( 4 n + 1 ) x 2 ∣ ∣ → x 2 as n → ∞
By the Ratio Test, the given series converges is x 2 < 1 x^2 < 1 x 2 < 1 x 2 > 1 x^2 > 1 x 2 > 1 x 2 = 1 x^2 = 1 x 2 = 1
∑ n = 1 ∞ 1 ⋅ 3 ⋅ 5 ⋅ … ⋅ ( 4 n − 3 ) 2 ⋅ 4 ⋅ 6 ⋅ … ⋅ ( 4 n − 2 ) ⋅ 1 4 n = ∑ n = 1 ∞ ( 4 n − 3 ) ! ! ( 4 n ) ! ! \sum_{n=1}^{\infty} \frac{1 \cdot 3 \cdot 5 \cdot \ldots \cdot (4n-3)}{2 \cdot 4 \cdot 6 \cdot \ldots \cdot (4n-2)} \cdot \frac{1}{4n} = \sum_{n=1}^{\infty} \frac{(4n-3)!!}{(4n)!!} n = 1 ∑ ∞ 2 ⋅ 4 ⋅ 6 ⋅ … ⋅ ( 4 n − 2 ) 1 ⋅ 3 ⋅ 5 ⋅ … ⋅ ( 4 n − 3 ) ⋅ 4 n 1 = n = 1 ∑ ∞ ( 4 n )!! ( 4 n − 3 )!!
The generating function for the Central Binomial Coefficients is
( 1 − 4 x ) − 1 / 2 = ∑ k = 0 ∞ ( 2 k k ) x k = 1 + 2 x + 6 x 2 + 20 x 3 + 70 x 4 + 252 x 5 + ⋯ (1 - 4x)^{-1/2} = \sum_{k=0}^{\infty} \binom{2k}{k} x^k = 1 + 2x + 6x^2 + 20x^3 + 70x^4 + 252x^5 + \cdots ( 1 − 4 x ) − 1/2 = k = 0 ∑ ∞ ( k 2 k ) x k = 1 + 2 x + 6 x 2 + 20 x 3 + 70 x 4 + 252 x 5 + ⋯
We can plug x = − 1 / 4 x = -1/4 x = − 1/4
( 1 − 4 ( − 1 4 ) ) − 1 / 2 = ∑ k = 0 ∞ ( 2 k k ) ( − 1 4 ) k = ∑ k = 0 ∞ ( − 1 ) k ( 2 k ) ! 4 k ( ( k ) ! ) 2 = = 1 + 2 ( − 1 4 ) + 6 ( − 1 4 ) 2 + 20 ( − 1 4 ) 3 + 70 ( − 1 4 ) 4 + 252 ( − 1 4 ) 5 + ⋯ = 1 + ∑ k = 1 ∞ ( − 1 ) k ( 2 k − 1 ) ! ! ( 2 k ) ! ! 2 2 = 1 + ∑ k = 1 ∞ ( − 1 ) k ( 2 k − 1 ) ! ! ( 2 k ) ! ! 1 − 2 2 = − ∑ k = 1 ∞ ( − 1 ) k ( 2 k − 1 ) ! ! ( 2 k ) ! ! \begin{array}{l}
\left(1 - 4\left(-\frac{1}{4}\right)\right)^{-1/2} = \sum_{k=0}^{\infty} \binom{2k}{k}\left(-\frac{1}{4}\right)^k = \sum_{k=0}^{\infty} (-1)^k \frac{(2k)!}{4^k \left((k)!\right)^2} = \\
= 1 + 2\left(-\frac{1}{4}\right) + 6\left(-\frac{1}{4}\right)^2 + 20\left(-\frac{1}{4}\right)^3 + 70\left(-\frac{1}{4}\right)^4 + 252\left(-\frac{1}{4}\right)^5 + \cdots \\
= 1 + \sum_{k=1}^{\infty} (-1)^k \frac{(2k-1)!!}{(2k)!!} \\
\frac{\sqrt{2}}{2} = 1 + \sum_{k=1}^{\infty} (-1)^k \frac{(2k-1)!!}{(2k)!!} \\
1 - \frac{\sqrt{2}}{2} = - \sum_{k=1}^{\infty} (-1)^k \frac{(2k-1)!!}{(2k)!!}
\end{array} ( 1 − 4 ( − 4 1 ) ) − 1/2 = ∑ k = 0 ∞ ( k 2 k ) ( − 4 1 ) k = ∑ k = 0 ∞ ( − 1 ) k 4 k ( ( k )! ) 2 ( 2 k )! = = 1 + 2 ( − 4 1 ) + 6 ( − 4 1 ) 2 + 20 ( − 4 1 ) 3 + 70 ( − 4 1 ) 4 + 252 ( − 4 1 ) 5 + ⋯ = 1 + ∑ k = 1 ∞ ( − 1 ) k ( 2 k )!! ( 2 k − 1 )!! 2 2 = 1 + ∑ k = 1 ∞ ( − 1 ) k ( 2 k )!! ( 2 k − 1 )!! 1 − 2 2 = − ∑ k = 1 ∞ ( − 1 ) k ( 2 k )!! ( 2 k − 1 )!! = 1 2 − 1 ⋅ 3 2 ⋅ 4 + 1 ⋅ 3 ⋅ 5 2 ⋅ 4 ⋅ 6 − 1 ⋅ 3 ⋅ 5 ⋅ 7 2 ⋅ 4 ⋅ 6 ⋅ 8 + 1 ⋅ 3 ⋅ 5 ⋅ 7 ⋅ 9 2 ⋅ 4 ⋅ 6 ⋅ 8 ⋅ 10 − 1 ⋅ 3 ⋅ 5 ⋅ 7 ⋅ 9 ⋅ 11 2 ⋅ 4 ⋅ 6 ⋅ 8 ⋅ 10 ⋅ 12 + … = 1 2 ⋅ 4 + 1 ⋅ 3 ⋅ 5 2 ⋅ 4 ⋅ 6 ⋅ 8 + 1 ⋅ 3 ⋅ 5 ⋅ 7 ⋅ 9 2 ⋅ 4 ⋅ 6 ⋅ 8 ⋅ 10 ⋅ 12 + ⋯ = ∑ n = 1 ∞ ( 4 n − 3 ) ! ! ( 4 n ) ! ! \begin{array}{l}
= \frac{1}{2} - \frac{1 \cdot 3}{2 \cdot 4} + \frac{1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6} - \frac{1 \cdot 3 \cdot 5 \cdot 7}{2 \cdot 4 \cdot 6 \cdot 8} + \frac{1 \cdot 3 \cdot 5 \cdot 7 \cdot 9}{2 \cdot 4 \cdot 6 \cdot 8 \cdot 10} - \frac{1 \cdot 3 \cdot 5 \cdot 7 \cdot 9 \cdot 11}{2 \cdot 4 \cdot 6 \cdot 8 \cdot 10 \cdot 12} + \dots \\
= \frac{1}{2 \cdot 4} + \frac{1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6 \cdot 8} + \frac{1 \cdot 3 \cdot 5 \cdot 7 \cdot 9}{2 \cdot 4 \cdot 6 \cdot 8 \cdot 10 \cdot 12} + \dots = \sum_{n=1}^{\infty} \frac{(4n - 3)!!}{(4n)!!} \\
\end{array} = 2 1 − 2 ⋅ 4 1 ⋅ 3 + 2 ⋅ 4 ⋅ 6 1 ⋅ 3 ⋅ 5 − 2 ⋅ 4 ⋅ 6 ⋅ 8 1 ⋅ 3 ⋅ 5 ⋅ 7 + 2 ⋅ 4 ⋅ 6 ⋅ 8 ⋅ 10 1 ⋅ 3 ⋅ 5 ⋅ 7 ⋅ 9 − 2 ⋅ 4 ⋅ 6 ⋅ 8 ⋅ 10 ⋅ 12 1 ⋅ 3 ⋅ 5 ⋅ 7 ⋅ 9 ⋅ 11 + … = 2 ⋅ 4 1 + 2 ⋅ 4 ⋅ 6 ⋅ 8 1 ⋅ 3 ⋅ 5 + 2 ⋅ 4 ⋅ 6 ⋅ 8 ⋅ 10 ⋅ 12 1 ⋅ 3 ⋅ 5 ⋅ 7 ⋅ 9 + ⋯ = ∑ n = 1 ∞ ( 4 n )!! ( 4 n − 3 )!!
The series
∑ n = 1 ∞ 1 ⋅ 3 ⋅ 5 ⋅ … ⋅ ( 4 n − 3 ) 2 ⋅ 4 ⋅ 6 ⋅ … ⋅ ( 4 n − 2 ) ⋅ x 2 n 4 n \sum_{n=1}^{\infty} \frac{1 \cdot 3 \cdot 5 \cdot \ldots \cdot (4n - 3)}{2 \cdot 4 \cdot 6 \cdot \ldots \cdot (4n - 2)} \cdot \frac{x^{2n}}{4n} n = 1 ∑ ∞ 2 ⋅ 4 ⋅ 6 ⋅ … ⋅ ( 4 n − 2 ) 1 ⋅ 3 ⋅ 5 ⋅ … ⋅ ( 4 n − 3 ) ⋅ 4 n x 2 n
is convergent for x ∈ [ − 1 , 1 ] x \in [-1, 1] x ∈ [ − 1 , 1 ]
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#340153 on Dec 2023