What is the sum from n=1 to infinity}(6e^{1/n of)/n ?

The sum from n=1 to infinity}(6e^{1/n of)/n is converges

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sum from n=1 to infinity}(6e^{1/n of)/n

n = 1 \sum \infty \frac{6 e ^{\frac{1}{n}}}{n}

converges

Solution steps

n = 1 \sum \infty \frac{6 e ^{\frac{1}{n}}}{n}

Apply the constant multiplication rule:

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

= 6 \cdot n = 1 \sum \infty \frac{e ^{\frac{1}{n}}}{n}

Apply Series Integral Test:

converges

= 6 converges

= converges

k = 1 \sum \infty 1

n = 4 \sum \infty \frac{1}{2 n ^{2} + n}

n = 1 \sum \infty \frac{n}{5 n - 3}

n = 0 \sum \infty \frac{2}{3 n}

n = 1 \sum \infty \frac{3}{n ^{\frac{5}{4}}}

What is the sum from n=1 to infinity}(6e^{1/n of)/n ?
The sum from n=1 to infinity}(6e^{1/n of)/n is converges