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

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

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

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

diverges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Series Divergence Test:

diverges

= 4 diverges

= diverges

n = 0 \sum \infty \frac{5}{2 n - 3}

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

n = 1 \sum \infty (- \frac{7}{8})^{n - 1}

n = 1 \sum \infty \frac{n + 1}{2 n - 1}

n = 0 \sum \infty (- 1)^{n - 1}

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