What is the sum from n=1 to infinity of (15)/n ?

The sum from n=1 to infinity of (15)/n is diverges

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sum from n=1 to infinity of (15)/n

n = 1 \sum \infty \frac{15}{n}

diverges

Solution steps

n = 1 \sum \infty \frac{15}{n}

Apply the constant multiplication rule:

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

= 15 \cdot n = 1 \sum \infty \frac{1}{n}

Apply Cauchy's Convergence Condition:

diverges

= 15 diverges

= diverges

\int_{0}^{1} x e^{2 x} d x

y^{^{'}} + 4 y = 6

- 6.25 t^{2} + 50 t

(x^{2} + 4 x y) d x + x d y = 0

x \to 1 lim (\frac{5}{( x - 1 ) ^{2}})

What is the sum from n=1 to infinity of (15)/n ?
The sum from n=1 to infinity of (15)/n is diverges