Question

Determine the convergence or divergence of the series. sum_{n=1}^infty(frac{1}{n^2}-frac{1}{n^3})

Series
ANSWERED
asked 2020-11-29
Determine the convergence or divergence of the series.
\(\sum_{n=1}^\infty(\frac{1}{n^2}-\frac{1}{n^3})\)

Answers (1)

2020-11-30
P-series test:
\(\sum_{n=1}^\infty\frac{1}{n^p}\)
If \(p>1\) series is convergent
If \(p\leq1\) 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
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