What is the sum from x=1 to infinity of (2^nx^n)/n ?

The sum from x=1 to infinity of (2^nx^n)/n is diverges

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sum from x=1 to infinity of (2^nx^n)/n

x = 1 \sum \infty \frac{2 ^{n} x ^{n}}{n}

diverges

Solution steps

n = 1 \sum \infty \frac{2 ^{n_{1}} n ^{n_{1}}}{n _{1}}

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

= \frac{2 ^{n}}{n} \cdot n_{1} = 1 \sum \infty n_{1}^{n}

Apply Series Divergence Test:

diverges

= \frac{2 ^{n}}{n} diverges

= diverges

n = 0 \sum \infty \frac{1000}{n}

n = 1 \sum \infty n^{k} e^{- n}

n = 0 \sum \infty \frac{e}{n}

n = 0 \sum \infty \frac{7}{π ^{n}}

n = 1 \sum \infty sin (3 n)

What is the sum from x=1 to infinity of (2^nx^n)/n ?
The sum from x=1 to infinity of (2^nx^n)/n is diverges