What is the sum from n=0 to infinity of ln(1/(n+1)) ?

The sum from n=0 to infinity of ln(1/(n+1)) is diverges

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sum from n=0 to infinity of ln(1/(n+1))

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

diverges

Solution steps

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

Apply log rule:

lo g_{a} (\frac{1}{x}) = - lo g_{a} (x)

= n = 0 \sum \infty - ln (n + 1)

Apply the constant multiplication rule:

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

= - n = 0 \sum \infty ln (n + 1)

Apply Series Divergence Test:

diverges

= - diverges

= diverges

\frac{d y}{d x} = \frac{2 x y - y ^{2}}{2 x y - x ^{2}}

y = 8 x sin (x)

x \to 0 lim (\frac{sin ( 3 ) ( x )}{x})

\int x x^{4} - 9 d x

\int_{0}^{3} (x^{2} - 2 x) d x

What is the sum from n=0 to infinity of ln(1/(n+1)) ?
The sum from n=0 to infinity of ln(1/(n+1)) is diverges