Question

use the integral test to determine whether the series is convergent or divergent.
$sum_{n=1}^{infty} n^2 e^{-n^3}$
evaluate the following integral.
$int_{1}^{infty} x^2 e^{-x^3} dx$
$\frac{1}{3e}$
since the integral is finite, the series is convergent.

1. - / 0.88 points

determine whether the series is convergent or divergent.
$1 + \frac{1}{2sqrt{2}} + \frac{1}{3sqrt{3}} + \frac{1}{4sqrt{4}} + \frac{1}{5sqrt{5}} dots$
the series is a convergent p - series with $p = $

1. - / 0.88 points

determine whether the series is convergent or divergent.
$sum_{n=1}^{infty} \frac{3}{n^2 + 25}$
$circ$ convergent
$circ$ divergent

Explanation:

Step1: Identify the general term of the series

The given series \(1+\frac{1}{2\sqrt{2}}+\frac{1}{3\sqrt{3}}+\frac{1}{4\sqrt{4}}+\frac{1}{5\sqrt{5}}+\cdots\) can be written in the general form \(\sum_{n = 1}^{\infty}\frac{1}{n\cdot n^{\frac{1}{2}}}=\sum_{n=1}^{\infty}\frac{1}{n^{1 + \frac{1}{2}}}=\sum_{n = 1}^{\infty}\frac{1}{n^{\frac{3}{2}}}\) using the rule of exponents \(a^m\cdot a^n=a^{m + n}\).

Step2: Recall the \(p -\)series test

The \(p -\)series is of the form \(\sum_{n=1}^{\infty}\frac{1}{n^{p}}\). The \(p -\)series \(\sum_{n = 1}^{\infty}\frac{1}{n^{p}}\) converges if \(p>1\) and diverges if \(p\leq1\).

Answer:

\(\frac{3}{2}\)