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Test the series of positive terms for convergence
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Asked by biaggi.max2002 | 31 Jan, 2022, 06:42: PM
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fraction numerator 1 over denominator 1 times 3 end fraction plus fraction numerator 2 over denominator 3 times 5 end fraction plus fraction numerator 3 over denominator 5 times 7 end fraction plus... equals fraction numerator straight n over denominator open parentheses 2 straight n minus 1 close parentheses open parentheses 2 straight n plus 1 close parentheses end fraction equals 1 fourth open parentheses fraction numerator 1 over denominator 2 straight n plus 1 end fraction plus fraction numerator 1 over denominator 2 straight n minus 1 end fraction close parentheses Consider comma limit as straight n rightwards arrow infinity of open parentheses straight a subscript straight n plus 1 end subscript over straight a subscript straight n close parentheses equals limit as straight n rightwards arrow infinity of fraction numerator open parentheses fraction numerator 1 over denominator 2 straight n plus 3 end fraction plus fraction numerator 1 over denominator 2 straight n plus 1 end fraction close parentheses over denominator begin display style fraction numerator 1 over denominator 2 straight n plus 1 end fraction end style plus begin display style fraction numerator 1 over denominator 2 straight n minus 1 end fraction end style end fraction equals limit as straight n rightwards arrow infinity of fraction numerator open parentheses 4 straight n plus 4 close parentheses open parentheses 2 straight n minus 1 close parentheses over denominator open parentheses 2 straight n plus 3 close parentheses 4 straight n end fraction equals limit as straight n rightwards arrow infinity of fraction numerator open parentheses 4 plus begin display style 4 over straight n end style close parentheses open parentheses 2 minus begin display style 1 over straight n end style close parentheses over denominator open parentheses 2 plus begin display style 3 over straight n end style close parentheses 4 end fraction equals fraction numerator open parentheses 4 plus 0 close parentheses open parentheses 2 minus 0 close parentheses over denominator open parentheses 2 plus 0 close parentheses 4 end fraction equals 1
When L<1, the series is convergent
When L>1, the seires is divergent.
Since, L=1, so we can't make any conclusion.
Thus, it is neither convergent nor divergent
Answered by Renu Varma | 03 Feb, 2022, 11:43: AM

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