What is the sum from n=1 to infinity of 4/(7^n)+3/n ?

The sum from n=1 to infinity of 4/(7^n)+3/n is diverges

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sum from n=1 to infinity of 4/(7^n)+3/n

n = 1 \sum \infty \frac{4}{7 ^{n}} + \frac{3}{n}

diverges

Solution steps

n = 1 \sum \infty \frac{4}{7 ^{n}} + \frac{3}{n}

Apply Series Comparison Test:

diverges

= diverges

(2 x y^{2} - 7) d x + (2 x^{2} y + 6) d y = 0

\frac{d}{d x} (ln (\frac{e ^{x}}{e ^{x} - 1}))

f (x) = x - 1, at x = 10

\frac{e ^{x}}{e ^{x} - e ^{- x}}

(\frac{d y}{d x}) + 2 y = e^{- x}

What is the sum from n=1 to infinity of 4/(7^n)+3/n ?
The sum from n=1 to infinity of 4/(7^n)+3/n is diverges