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

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

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

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

diverges

Solution steps

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

2 n = 2 n

= 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

\int \frac{( x ^{2} + 2 )}{( x + 2 )} d x

x \to 0 lim (6 x + ∣ x - 5 ∣)

\frac{d}{d x} (\frac{2 x ^{2} + 2 x + 8}{x})

f (x) = lo g_{2} (x)

x \to - 3 - lim (f (x))

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