What is the sum from n=1 to infinity of 6/(n+5) ?

The sum from n=1 to infinity of 6/(n+5) is diverges

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sum from n=1 to infinity of 6/(n+5)

n = 1 \sum \infty \frac{6}{n + 5}

diverges

Solution steps

n = 1 \sum \infty \frac{6}{n + 5}

Apply the constant multiplication rule:

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

= 6 \cdot n = 1 \sum \infty \frac{1}{n + 5}

Apply Series Integral Test:

diverges

= 6 diverges

= diverges

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

n = 0 \sum \infty \frac{π}{2}

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

n = 0 \sum \infty ln (9 - n)

n = 1 \sum \infty 1 + n

What is the sum from n=1 to infinity of 6/(n+5) ?
The sum from n=1 to infinity of 6/(n+5) is diverges