What is the sum from n=1 to infinity}((2^{2n of n!))/(2n+1) ?

The sum from n=1 to infinity}((2^{2n of n!))/(2n+1) is diverges

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sum from n=1 to infinity}((2^{2n of n!))/(2n+1)

n = 1 \sum \infty \frac{( 2 ^{2 n} n ! )}{2 n + 1}

diverges

Solution steps

n = 1 \sum \infty \frac{2 ^{2 n} n !}{2 n + 1}

2^{2 n} = 4^{n}

= n = 1 \sum \infty \frac{4 ^{n} n !}{2 n + 1}

Apply Series Ratio Test:

diverges

= diverges

e^{- 3 t} \cdot sin (t)

x \to 3 + lim (\frac{2}{x + 3})

F (x) = x^{4} - 8 x^{2} + 2

x \to 2 lim (n x + 2)

\frac{d}{d x} (sin (5) x^{5})

What is the sum from n=1 to infinity}((2^{2n of n!))/(2n+1) ?
The sum from n=1 to infinity}((2^{2n of n!))/(2n+1) is diverges