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

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

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

n = 1 \sum \infty \frac{4 ^{n}}{n + 1}

diverges

Solution steps

n = 1 \sum \infty \frac{4 ^{n}}{n + 1}

Apply Series Ratio Test:

diverges

= diverges

n = 1 \sum \infty \frac{1}{n ( n + 8 )}

168 (2 x + 5)^{5}

\frac{d}{d x} (lo g_{7} (x^{2} - 7))

f (x) = 4 x^{4} - 3 x^{\frac{5}{2}} + 2

\frac{\partial}{\partial x} (3 e^{y} + x + x^{3} + 6)

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