What is the sum from n=1 to infinity of 1/(10ln(n)) ?

The sum from n=1 to infinity of 1/(10ln(n)) is diverges

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sum from n=1 to infinity of 1/(10ln(n))

n = 1 \sum \infty \frac{1}{10 ln ( n )}

diverges

Solution steps

n = 2 \sum \infty \frac{1}{10 ln ( n )}

Apply the constant multiplication rule:

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

= \frac{1}{10} \cdot n = 2 \sum \infty \frac{1}{ln ( n )}

Apply Series Integral Test:

diverges

= \frac{1}{10} diverges

= diverges

n = 0 \sum \infty \frac{( - 5 ) ^{n}}{n}

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

n = 0 \sum \infty ln (3 - n)

n = 0 \sum \infty \frac{1}{3 ^{n} - 1}

n = 0 \sum \infty 0.52 (0.01)^{n}

What is the sum from n=1 to infinity of 1/(10ln(n)) ?
The sum from n=1 to infinity of 1/(10ln(n)) is diverges