What is the sum from n=0 to infinity of 4/(2n+1) ?

The sum from n=0 to infinity of 4/(2n+1) is diverges

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sum from n=0 to infinity of 4/(2n+1)

n = 0 \sum \infty \frac{4}{2 n + 1}

diverges

Solution steps

n = 0 \sum \infty \frac{4}{2 n + 1}

Apply the constant multiplication rule:

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

= 4 \cdot n = 0 \sum \infty \frac{1}{2 n + 1}

Apply Series Integral Test:

diverges

= 4 diverges

= diverges

f (x) = 6 x e^{x}, (0, 0)

y = tan (2 x)

\frac{\partial}{\partial x} (- 3 sin (3 x))

4 x - \frac{x ^{3}}{4}

\int - y sin (y) d y

What is the sum from n=0 to infinity of 4/(2n+1) ?
The sum from n=0 to infinity of 4/(2n+1) is diverges