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The series 
  • a)
    conditional convergent
  • b)
    absolutely convergent
  • c)
    divergent
  • d)
    None of the above
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
The seriesa)conditional convergentb)absolutely convergentc)divergentd)...
Let the given series be denoted by ∑μn, then
∑|μn|

Compare this series ∑vn with the auxiliary series,

Then,

which is finite quantity.
Hence, ∑un and ∑wn are either both convergent or both divergent. But ∑wn 

Hence, the series ∑vn is divergent.
Also, in the series ∑un, we find that its term are alternately positive and negative, its terms are continuously decreasing and

Thus, all condition of Leibnitz’s test are satisfied and as such ∑un is convergent. H ence, the given series ∑un is conditionally convergent.
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Most Upvoted Answer
The seriesa)conditional convergentb)absolutely convergentc)divergentd)...
Here ,the given series is of positive terms and therefore the series will be Divergent.
so, option (c) is correct
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Community Answer
The seriesa)conditional convergentb)absolutely convergentc)divergentd)...
In the above we have used the result that 1/n → 0 as n → ∞ and a result about combining convergent sequences and noting that the denominator converges to a non-zero value.
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The seriesa)conditional convergentb)absolutely convergentc)divergentd)None of the aboveCorrect answer is option 'A'. Can you explain this answer?
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