What is the sum from n=0 to infinity of 8ln(4-n) ?

The sum from n=0 to infinity of 8ln(4-n) is diverges

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sum from n=0 to infinity of 8ln(4-n)

n = 0 \sum \infty 8 ln (4 - n)

diverges

Solution steps

n = 5 \sum \infty 8 ln (4 - n)

Apply the constant multiplication rule:

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

= 8 \cdot n = 5 \sum \infty ln (4 - n)

Apply Series Divergence Test:

diverges

= 8 diverges

= diverges

n = 1 \sum \infty (- \frac{5}{6})^{n - 1}

n = 2 \sum \infty (\frac{3}{8})^{3 n}

n = 0 \sum \infty (1 + \frac{6}{n})^{n}

n = 1 \sum \infty 9 (0.9)^{n - 1}

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

What is the sum from n=0 to infinity of 8ln(4-n) ?
The sum from n=0 to infinity of 8ln(4-n) is diverges