Proving that ∑∞n=1(−1)n+1n+12n∑∞n=1(−1)n+1n+12n\sum_{n=1}^\infty (-1)^{n+1} \frac{n+1}{2n} diverges.

I want to prove that ∑∞n=1(−1)n+1n+12n\sum_{n=1}^\infty (-1)^{n+1} \frac{n+1}{2n} diverges but am having trouble. My idea is to find a subsequence of the sequence of partial sums (Sn)(S_n) that diverges. I have (Sn)(S_n) as (1+∑nk=1(−1)k2+k2+2k)(1 + \sum_{k=1}^n (-1)^k \frac{2+k}{2+2k}) but am still stuck. I would appreciate any help.

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A necessary condition for convergence of an infinite series is that the general term approach zero.
– bof
2 days ago

  

 

Right, I tried proving that the sequence doesn’t converge to 0 but can’t work out the details.
– CuriousKid7
2 days ago

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The absolute value of the nthn^{\text{th}} term is n+12n>n2n=12.\frac{n+1}{2n}\gt\frac n{2n}=\frac12. Note that an→0a_n\to0 implies |an|→0.|a_n|\to0. If |an|≥12|a_n|\ge\frac12 for all nn then |an||a_n| does not approach 00 and neither does an.a_n.
– bof
2 days ago

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2 Answers
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All you need to do in this case is note that limn→∞(n+1)(2n)=12\lim_{n \to \infty}{\frac{(n+1)}{(2n)}}=\frac{1}{2} which means that since the sequence doesn’t approach 00, the alternating series itself must, as a result, diverge.

  

 

It seems like you’re using the converse of the alternating series test to prove divergence—which you can’t do.
– CuriousKid7
2 days ago

  

 

It’s called the divergence test. You can do it, even for non-alternating series.
– ziggurism
2 days ago

  

 

@zuggurism Yes, but the sequence of terms in the series is not ((n+1) / (2n)) but rather ((−1)^n+1 (n+1)/2n). So this is the sequence I need to show doesn’t converge to 0.
– CuriousKid7
2 days ago

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@CuriousKid7 Theorem. limn→∞an=0⟺limn→∞|an|=0.\lim_{n\to\infty}a_n=0\iff\lim_{n\to\infty}|a_n|=0.
– bof
2 days ago

Considering an=(−1)n+1(n+1)2na_n=(-1)^{n+1}\frac{ (n+1)}{2 n} use the ratio test an+1an=−n(n+2)(n+1)2⟹|an+1an|∼1−1n2\frac{a_{n+1}}{a_n}=-\frac{n (n+2)}{(n+1)^2}\implies \left|\frac{a_{n+1}}{a_n}\right|\sim 1-\frac 1 {n^2} which show that it is inconclusive.

Now, use Raabe’s test n(|anan+1|−1)=1n+2<1n \left(\left|\frac{a_n}{a_{n+1}}\right|-1\right)=\frac{1}{n+2}<1 then divergence.      With Raabe's test you can only decide whether a series converges absolutely. – Dominik 2 days ago