What is the sum from n=1 to infinity of (2n+1)/(3n) ?

The sum from n=1 to infinity of (2n+1)/(3n) is diverges

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sum from n=1 to infinity of (2n+1)/(3n)

n = 1 \sum \infty \frac{2 n + 1}{3 n}

diverges

Solution steps

n = 1 \sum \infty \frac{2 n + 1}{3 n}

Apply the constant multiplication rule:

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

= \frac{1}{3} \cdot n = 1 \sum \infty \frac{2 n + 1}{n}

Apply Series Divergence Test:

diverges

= \frac{1}{3} diverges

= diverges

n = 1 \sum \infty 3 (0.6)^{n - 1}

n = 1 \sum \infty \frac{2 ^{n}}{7 ^{n}}

n = 1 \sum \infty \frac{2}{6 ^{n}}

n = 1 \sum \infty n^{3} + 1

k = 3 \sum \infty \frac{1}{( k - 2 ) ^{5}}

What is the sum from n=1 to infinity of (2n+1)/(3n) ?
The sum from n=1 to infinity of (2n+1)/(3n) is diverges