Determine the convergence or divergence of the series. ∑n=1∞(1n2−1n3)

P-series test: ∑n=1∞1np If p&gt;1 series is convergent If p≤1 series is divergent Series sum or difference rule: If ∑n=1∞an and ∑n=1∞bn are both convergent then ∑n=1∞(an±bn)=∑n=1∞an±∑n=1∞bn And sum of two convergent series is also convergent Given that ∑n=1∞1np Let ∑n=1∞an=∑n=1∞1n2 and ∑n=1∞bn=∑n=1∞1n3 Using p-series test both ∑an and ∑bn are convergent so using series sum or difference rule ∑n=1∞(1n2−1n3)=∑n=1∞1n2−∑n=1∞1n3 Since ∑n=1∞an=1n2 and ∑n=1∞bn=1n3 both are convergent. Hence ∑n=1∞(1n2−1n3) is also convergent

Answered: "Determine the convergence or divergence of the series...."

bobbie71G

Answered question

2020-11-29

Determine the convergence or divergence of the series.

\sum_{n = 1}^{\infty} (\frac{1}{n^{2}} - \frac{1}{n^{3}})

Answer & Explanation

faldduE

Skilled2020-11-30Added 109 answers

P-series test:

\sum_{n = 1}^{\infty} \frac{1}{n^{p}}

p > 1

series is convergent
If

p \leq 1

series is divergent
Series sum or difference rule:
If

\sum_{n = 1}^{\infty} a_{n}

and

\sum_{n = 1}^{\infty} b_{n}

are both convergent then

\sum_{n = 1}^{\infty} (a_{n} \pm b_{n}) = \sum_{n = 1}^{\infty} a_{n} \pm \sum_{n = 1}^{\infty} b_{n}

And sum of two convergent series is also convergent
Given that

\sum_{n = 1}^{\infty} \frac{1}{n^{p}}

Let

\sum_{n = 1}^{\infty} a_{n} = \sum_{n = 1}^{\infty} \frac{1}{n^{2}}

and

\sum_{n = 1}^{\infty} b_{n} = \sum_{n = 1}^{\infty} \frac{1}{n^{3}}

Using p-series test both

\sum a_{n}

and

\sum b_{n}

are convergent so using series sum or difference rule

\sum_{n = 1}^{\infty} (\frac{1}{n^{2}} - \frac{1}{n^{3}}) = \sum_{n = 1}^{\infty} \frac{1}{n^{2}} - \sum_{n = 1}^{\infty} \frac{1}{n^{3}}

Since

\sum_{n = 1}^{\infty} a_{n} = \frac{1}{n^{2}}

and

\sum_{n = 1}^{\infty} b_{n} = \frac{1}{n^{3}}

both are convergent. Hence

\sum_{n = 1}^{\infty} (\frac{1}{n^{2}} - \frac{1}{n^{3}})

is also convergent

Jeffrey Jordon

Expert2021-12-27Added 2605 answers

Answer is given below (on video)

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