I want to check, whether ∞∑n=11+(−1)nn converges or diverges.
Leibniz's test failed, and ratio test just made it even more complicated, so i tried to use the comparison test, but i can't find a suitable series so that limn→∞anbn exists..
Answer
To prove: ∑n≥11+(−1)nn diverges.
Proof: ∑n≥11+(−1)nn=∑k≥11+(−1)2k2k+∑k≥11+(−1)2k−12k−1=∑k≥122k+∑k≥102k−1 =∑k≥11k
Because ∑k≥11k diverges, ∑n≥11+(−1)nn diverges as well. \qed
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