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I am not sure if the title I have given is appropriate. Please edit it if you find necessary.

Can someone please help me about the convergence of this series $\frac{1}{4}-1+\frac{3}{4}+\frac{5}{4}-3+\frac{7}{4}+\frac{9}{4}-5+\frac{11}{4}+\cdots$.

What I tried was the following:

Leaving the first term, let $\sum a_n$ be the given series. Then we see that \begin{align*} a_{3n-2}=&-(2n-1)\\ a_{3n-1}=&\frac{1}{4}(4n-1)\\ a_{3n}=&\frac{1}{4}(4n+1) \end{align*} so that $\frac{a_{3n}}{a_{3n-1}}\rightarrow 1$ and $\frac{a_{3n-1}}{a_{3n-2}}\rightarrow -\frac{1}{2}$ as $n\rightarrow \infty$.

What next ?

Please help me.

KON3
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1 Answers1

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Let $S_N=\sum_{n=0}^N a_n$ then it is easy to see that $S_{3N}=0$, $S_{3N+1}=a_{3N+1}=-(2N+1)$ (or one can take $S_{3N-1}=-a_{3N}=N+\frac{1}{4}$). So the sequence $\{S_n\}_{n\geq 0}$ has one subsequence which goes to $0$ and another one that goes to $-\infty$. Hence $S_N$ can not be convergent.

P.S. A more direct way: the generic term $a_n$ is not infinitesimal...

Robert Z
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  • Thank you so much. I have got the answer – KON3 Aug 11 '16 at 11:09
  • Assuming that the general terms are $1/4$, $-1$, $3/4$, $5/4$ and so on, which isn't directly stated in the question (although it is in the author's partial answer). – Santiago Aug 11 '16 at 11:22