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

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

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sum from n=1 to infinity of in(1/n)

n = 1 \sum \infty in (\frac{1}{n})

diverges

Solution steps

i = 1 \sum \infty ni \frac{1}{i}

Apply the constant multiplication rule:

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

= i \cdot n = 1 \sum \infty n \frac{1}{n}

n \frac{1}{n} = 1

= i \cdot n = 1 \sum \infty 1

Every infinite sum of a non-zero constant diverges

= i diverges

= diverges

\frac{d}{d x} (\frac{3 x - 2}{2 x + 3})

g (t) = t^{7} cos (t)

f (x) = (2 x - 3)^{4}

\int 6 y (x - y) d y

x \to 0 lim (\frac{2 x + 4 - 2}{x + 1 - 1})

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