What is the sum from n=3 to infinity of-5/(n^{1.01)} ?

The sum from n=3 to infinity of-5/(n^{1.01)} is converges

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sum from n=3 to infinity of-5/(n^{1.01)}

n = 3 \sum \infty - \frac{5}{n ^{1.01}}

converges

Solution steps

n = 3 \sum \infty - \frac{5}{n ^{1.01}}

Apply the constant multiplication rule:

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

= - 5 \cdot n = 3 \sum \infty \frac{1}{n ^{1.01}}

Apply p-Series Test:

converges

= - 5 converges

= converges

f (x) = 7 x^{3} - 5 x + 8

cos (3 t)

\frac{d}{d x} (\frac{10 x}{( x ^{2} + 1 ) ^{2}})

\frac{d y}{d t} = (t + y)^{2} - 1

x \to 5 lim (9 - x^{2})

What is the sum from n=3 to infinity of-5/(n^{1.01)} ?
The sum from n=3 to infinity of-5/(n^{1.01)} is converges