infinite series - 8

[logistic_guy]

\(\displaystyle \textcolor{indigo}{\bold{Solve.}}\)

\(\displaystyle \sum_{k=1}^{\infty}k^{-9/10}\)

 

[khansaheb]

\(\displaystyle \textcolor{indigo}{\bold{Solve.}}\)

\(\displaystyle \sum_{k=1}^{\infty}k^{-9/10}\)

Please show us what you have tried and exactly where you are stuck.

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[logistic_guy]

\(\displaystyle \textcolor{indigo}{\bold{Solve.}}\)

\(\displaystyle \sum_{k=1}^{\infty}k^{-9/10}\)

This problem can be solved in many different ways. Of course in my case, I am interested in finding the convergence or divergence of the series.

\(\displaystyle \sum_{k=1}^{\infty}k^{-9/10} = \sum_{k=1}^{\infty}\frac{1}{k^{9/10}}\)

The series that has this structure:

\(\displaystyle \sum_{k=1}^{\infty}\frac{1}{k^{p}}\)

usually can be solved by the integral test.

Let us invent a new test that will help us answer this type of series quickly in the future.

We will call it the \(\displaystyle \bold{p}\)-\(\displaystyle \bold{series}\) test.

If we use the integral test, we get:

\(\displaystyle \int_{1}^{\infty}\frac{1}{x^p} \ dx = \frac{x^{1 - p}}{1 - p}\bigg|_{1}^{\infty}\)

Obviously \(\displaystyle p\) cannot be \(\displaystyle 1\).

Test some values in the solution of this integral and you will find out that it converges only when \(\displaystyle p > 1\).

In other words any series with this structure:

\(\displaystyle \sum_{k=1}^{\infty}\frac{1}{k^{p}} \longrightarrow\) diverges when \(\displaystyle p \leq 1\).

Then,

\(\displaystyle \sum_{k=1}^{\infty}k^{-9/10} = \sum_{k=1}^{\infty}\frac{1}{k^{9/10}}\)

diverges by the \(\displaystyle \bold{p}\)-\(\displaystyle \bold{series}\) test since \(\displaystyle \frac{9}{10} \leq 1\)