What is the sum from n=1 to infinity of 1/((2n-1)2n) ?

The sum from n=1 to infinity of 1/((2n-1)2n) is converges

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sum from n=1 to infinity of 1/((2n-1)2n)

n = 1 \sum \infty \frac{1}{( 2 n - 1 ) 2 n}

converges

Solution steps

n = 1 \sum \infty \frac{1}{( 2 n - 1 ) \cdot 2 n}

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

= \frac{1}{2} \cdot n = 1 \sum \infty \frac{1}{( 2 n - 1 ) n}

Apply Series Integral Test:

converges

= \frac{1}{2} converges

= converges

h \to 0 lim (4 (h + 2))

y^{^{'}} + 3 y = 1

\frac{d y}{d x} = x^{2} y^{3}

y^{^{''}} + 6 y^{^{'}} + 9 y = 3 x + 5, y (0) = 1, y^{^{'}} (0) = - 1

\int_{0}^{1} (x^{3} + x) d x

What is the sum from n=1 to infinity of 1/((2n-1)2n) ?
The sum from n=1 to infinity of 1/((2n-1)2n) is converges