I try to study the convergence or divergence of the series∑n=2∞(−1)nn+(−1)n. We can see that this series diverges absolutely, because ∑n=2∞1|n+(−1)n| is a rearrangement of ∑n=21n which equals to +∞. Consequently, we can't say for sure that the rearrangement ∑n=2∞(−1)nn+(−1)n of ∑n=2∞(−1)nn converges. That's the part where I am stuck. How exactly can I study the convergence (or not) of the first series?Should I apply Cauchy's criterion and prove thatsn=∑k=2n(−1)kk+(−1)k,n∈N is a basic sequence?

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I try to study the convergence or divergence of the series∑n=2∞(−1)nn+(−1)n. We can see that this series diverges absolutely, because ∑n=2∞1|n+(−1)n| is a rearrangement of ∑n=21n which equals to +∞. Consequently, we can&#039;t say for sure that the rearrangement ∑n=2∞(−1)nn+(−1)n of ∑n=2∞(−1)nn converges. That&#039;s the part where I am stuck. How exactly can I study the convergence (or not) of the first series?Should I apply Cauchy&#039;s criterion and prove thatsn=∑k=2n(−1)kk+(−1)k,n∈N is a basic sequence?

levurdondishav4 · Accepted Answer

Without giving a formal proof of it (or even a formal statement of the theorem), the following is true:You can do any bounded amount of rearranging of an infinite series without affecting its (conditional) convergence/divergence.For the series at hand, note that∑n=2∞(−1)nn+(−1)n=−13+12+−15+14+−17+16+⋯=12−13+14−15+16−17+⋯.where the &#039;&#039;bounded&#039;&#039; rearranging is to swap pairs of adjacent terms. The rearranged sequence now meets all the criteria of Leibniz&#039;s test, so the series is conditionally convergent. (The rearrangement makes it easy to see that the sum is 1−ln⁡2.)A somewhat more formal definition of &#039;&#039;bounded&#039;&#039; rearrangement is that no term is the rearranged series is ever more than B positions from where it appears in the original series, for some fixed positive number B. (In this example, B=1.) Somewhat more formally, a rearrangement ∑aπ(n) or ∑an, where π:N→N is a bijection (aka permutation), is &#039;&#039;bounded&#039;&#039; (by B) if |π(n)−n|≤B for all n.

kalupunangh · Accepted Answer

Note that  ∑n=2m(−1)nn+(−1)n=∑n=2m(−1)nn+(−1)nn−(−1)nn−(−1)n=∑n=2m(−1)nnn2−1−∑n=2m1n2−1,  and nn2−1→0 monotonically. Try to supply the details in showing that both series on the RHS are convergent.

alenahelenash · Accepted Answer

Here is a slightly different approach: For large n, we kind of expect (−1)nn+(−1)n to behave similarly as (−1)nn, whose sum we know how to control. In light of this, we study the difference: ∑n=2N((−1)nn+(−1)n−(−1)nn)=−∑n=2N1(n+(−1)n)n Now it is not hard to verify that the partial sum on the right-hand side converges absolutely as N→∞. Either the direct comparison or limit comparison will work.

I try to study the convergence or divergence of the series \sum_{n=2}^{\infty} \frac{(-1)^{n}

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