What is the sum from n=0 to infinity of 1/(2(2n+1)) ?

The sum from n=0 to infinity of 1/(2(2n+1)) is diverges

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sum from n=0 to infinity of 1/(2(2n+1))

n = 0 \sum \infty \frac{1}{2 ( 2 n + 1 )}

diverges

Solution steps

n = 0 \sum \infty \frac{1}{2 ( 2 n + 1 )}

Apply the constant multiplication rule:

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

= \frac{1}{2} \cdot n = 0 \sum \infty \frac{1}{2 n + 1}

Apply Series Integral Test:

diverges

= \frac{1}{2} diverges

= diverges

\int (x^{7} - 2 x^{6})^{4} (7 x^{6} - 12 x^{5}) d x

x \to 0 lim (\frac{50 x ^{2}}{sin ( x ) + 50 x ^{2}})

\int_{0}^{1} e^{2 x} d x

f (x) = x^{3} - 2 x - 2, at x = - 1

f (t) = \frac{9}{t ^{2}}

What is the sum from n=0 to infinity of 1/(2(2n+1)) ?
The sum from n=0 to infinity of 1/(2(2n+1)) is diverges