What is the sum from n=1 to infinity of (3n)/(n+1) ?

The sum from n=1 to infinity of (3n)/(n+1) is diverges

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sum from n=1 to infinity of (3n)/(n+1)

n = 1 \sum \infty \frac{3 n}{n + 1}

diverges

Solution steps

n = 1 \sum \infty \frac{3 n}{n + 1}

Apply the constant multiplication rule:

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

= 3 \cdot n = 1 \sum \infty \frac{n}{n + 1}

Apply Series Divergence Test:

diverges

= 3 diverges

= diverges

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

\frac{d}{d x} (x^{2 s i n (x)})

f (x) = 2 π x^{2}

f (x) = 2 x^{2} ln (2 x)

\frac{\partial}{\partial x} ((ln (x^{2} + y^{2}))^{\frac{1}{2}})

What is the sum from n=1 to infinity of (3n)/(n+1) ?
The sum from n=1 to infinity of (3n)/(n+1) is diverges