What is the sum from n=1 to infinity of ne^{3n} ?

The sum from n=1 to infinity of ne^{3n} is diverges

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sum from n=1 to infinity of ne^{3n}

n = 1 \sum \infty n e^{3 n}

diverges

Solution steps

n = 1 \sum \infty n e^{3 n}

Apply Series Ratio Test:

diverges

= diverges

x \to 0 + lim (\frac{6}{x ^{2}})

\int \frac{x}{x ^{2} + 2} d x

4 x, 64 - x^{2}, [- 6.928, 0]

\int 2 x sin^{4} (x^{2} - 1) d x

x \to 0 lim (2 x^{2} + 5)

What is the sum from n=1 to infinity of ne^{3n} ?
The sum from n=1 to infinity of ne^{3n} is diverges