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Interactive · 5 questions

A series is absolutely convergent if the series consisting of the absolute values of the terms is convergent. More formally, ∑i=1∞ai\sum_{i=1}^\infty a_i is absolutely convergent iff ∑i=1∞∣ai∣.\sum_{i=1}^\infty |a_i|.

Is the series ∑i=1∞(−1)ii\sum_{i=1}^{\infty} \frac{(-1)^i}{i} convergent?

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Interactive · 1 question

The series ∑an\sum a_n is said to be absolutely convergent if the series ∑∣an∣\sum|a_n| is convergent.

Notice that absolute convergence implies convergence. Suppose ∑an\sum a_n is absolutely convergent. Since (by the definition of absolute value) 0≤an+∣an∣≤2∣an∣,0\leq a_n + |a_n|\leq 2|a_n|, by the series comparison test we know that ∑an+∣an∣\sum a_n + |a_n| is convergent.

By definition of absolute convergence, ∑∣an∣\sum |a_n| is convergent. Subtracting shows that ∑(an+∣an∣)−∣an∣=∑an\sum (a_n+|a_n|) - |a_n| = \sum a_n is convergent, as desired.

Does the series ∑n=2∞(−1)⌊n2n−1⌋⋅1n3\sum_{n=2}^{\infty}(-1)^{\lfloor \frac{n^2}{n-1}\rfloor}\cdot\frac{1}{n^3} converge?

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Type a concept, like “slope” or “photosynthesis”.

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