What is the sum from n=1 to infinity of 5/(2n+1) ?

The sum from n=1 to infinity of 5/(2n+1) is diverges

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sum from n=1 to infinity of 5/(2n+1)

n = 1 \sum \infty \frac{5}{2 n + 1}

diverges

Solution steps

n = 1 \sum \infty \frac{5}{2 n + 1}

Apply the constant multiplication rule:

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

= 5 \cdot n = 1 \sum \infty \frac{1}{2 n + 1}

Apply Series Integral Test:

diverges

= 5 diverges

= diverges

n = 1 \sum \infty 3 n e^{- n}

n = 0 \sum \infty \frac{e ^{- 2 n}}{1}

n = 1 \sum \infty \frac{1}{n ln ^{3} ( n )}

n = 1 \sum \infty \frac{1}{n ^{2.3}}

n = 1 \sum \infty \frac{1}{4 ( n + 1 )}

What is the sum from n=1 to infinity of 5/(2n+1) ?
The sum from n=1 to infinity of 5/(2n+1) is diverges