What is the sum from n=1 to infinity of 1/(2sqrt(n)) ?

The sum from n=1 to infinity of 1/(2sqrt(n)) is diverges

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sum from n=1 to infinity of 1/(2sqrt(n))

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

diverges

Solution steps

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

Apply the constant multiplication rule:

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

= \frac{1}{2} \cdot n = 1 \sum \infty \frac{1}{n}

Apply p-Series Test:

diverges

= \frac{1}{2} diverges

= diverges

x = 0 \sum \infty \frac{1}{5 ^{x}}

n = 1 \sum \infty \frac{1}{1. 1 ^{n}}

k = 1 \sum \infty \frac{1}{16 k ^{2} - 4}

n = 1 \sum \infty \frac{n}{n ^{2} - 2 n + 5}

n = 1 \sum \infty \frac{7}{1 + n ^{2}}

What is the sum from n=1 to infinity of 1/(2sqrt(n)) ?
The sum from n=1 to infinity of 1/(2sqrt(n)) is diverges