What is the sum from n=0 to infinity of (6^n)/(n+1) ?

The sum from n=0 to infinity of (6^n)/(n+1) is diverges

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sum from n=0 to infinity of (6^n)/(n+1)

n = 0 \sum \infty \frac{6 ^{n}}{n + 1}

diverges

Solution steps

n = 0 \sum \infty \frac{6 ^{n}}{n + 1}

Apply Series Ratio Test:

diverges

= diverges

x \to - \frac{7 π}{2} + lim (sec (x))

\int (x^{2} + 7) \cdot ln (x) d x

x \to 0 lim (\frac{1}{cos ( 19 x )})

\frac{d}{d x} (6 - x^{2})

\int (2 x + 1) (2 x - 1) d x

What is the sum from n=0 to infinity of (6^n)/(n+1) ?
The sum from n=0 to infinity of (6^n)/(n+1) is diverges