What is the sum from n=1 to infinity of n+(7/8)^n ?

The sum from n=1 to infinity of n+(7/8)^n is diverges

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sum from n=1 to infinity of n+(7/8)^n

n = 1 \sum \infty n + (\frac{7}{8})^{n}

diverges

Solution steps

n = 1 \sum \infty n + (\frac{7}{8})^{n}

Apply Series Divergence Test:

diverges

= diverges

y = x^{3} - x^{2} + x - 1, (1, 0)

\int x^{4} ln^{2} (x) d x

\frac{d}{d x} (\frac{4 x}{x + 4})

y = 8 + 5 x^{2} - 2 x^{3}

\frac{4 s}{( s + 2 ) ^{2}}

What is the sum from n=1 to infinity of n+(7/8)^n ?
The sum from n=1 to infinity of n+(7/8)^n is diverges