What is the sum from n=1 to infinity of 5nsin(4/n) ?

The sum from n=1 to infinity of 5nsin(4/n) is diverges

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sum from n=1 to infinity of 5nsin(4/n)

n = 1 \sum \infty 5 n sin (\frac{4}{n})

diverges

Solution steps

n = 1 \sum \infty 5 n sin (\frac{4}{n})

Apply the constant multiplication rule:

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

= 5 \cdot n = 1 \sum \infty n sin (\frac{4}{n})

Apply Series Divergence Test:

diverges

= 5 diverges

= diverges

k = 1 \sum \infty k e^{- 2 k^{2}}

k = 1 \sum \infty \frac{9 k ^{3}}{k !}

n = 0 \sum \infty \frac{1}{3 n ^{2} + 8}

n = 0 \sum \infty n \cdot 0. 5^{n}

n = 1 \sum \infty \frac{1}{∣ n ∣}

What is the sum from n=1 to infinity of 5nsin(4/n) ?
The sum from n=1 to infinity of 5nsin(4/n) is diverges