What is the sum from n=1 to infinity of (3n)/(4n+1) ?

The sum from n=1 to infinity of (3n)/(4n+1) is diverges

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sum from n=1 to infinity of (3n)/(4n+1)

n = 1 \sum \infty \frac{3 n}{4 n + 1}

diverges

Solution steps

n = 1 \sum \infty \frac{3 n}{4 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}{4 n + 1}

Apply Series Divergence Test:

diverges

= 3 diverges

= diverges

\frac{( x + 2 ) ^{(2)} ( d y )}{( d x )} = 5 - 8 y - 4 x y

\int \frac{( y - 2 )}{y + 3} d y

\int_{0}^{0.6} 2 x e^{- x} d x

y^{^{'}} = 2 y

\frac{d y}{d x} = y^{\frac{1}{3}}

What is the sum from n=1 to infinity of (3n)/(4n+1) ?
The sum from n=1 to infinity of (3n)/(4n+1) is diverges