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I was thinking of using some sort of an alternating sequence for $(a_{n})_{n=1}$, but wasn't sure where to go. I know that an alternating series is not divergent, it's just not convergent since it doesn't tend to infinity but rather oscillates between two numbers.

Any ideas are appreciated.

Link to previously asked question: Sequences and series and the alternating series test

If the solutions are correct, can someone explain them properly please.

  • Why not let $a_n$ be the terms of a convergent nonalternating series and render $q_n=1/a_n$? Am I missing something subtle in the language here? – Oscar Lanzi Mar 25 '19 at 10:07
  • @OscarLanzi $q_n$ is given to you, you're not free to define it. – Jürgen Sukumaran Mar 25 '19 at 10:07
  • You could define $a_n = 1_n/q_n$ where $1_n$ is either 0 or 1; $1_n=1$ only when $q_n>0$ and $q_n$ is "sufficiently large;" $1_n=1$ for an infinite number of indices $n$. It remains to define a precise condition for "sufficiently large." – Michael Mar 25 '19 at 10:29

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We can find integers $n_1<n_2<\cdots$ such that $q_{n_k} >2^{k}$. Let $a_n=0$ if $n \notin \{n_1,n_2,\cdots\}$ and $a_{n_k}=q_{n_k}^{-1}$. $\sum a_n =\sum a_{n_k} <\sum \frac 1 {2^{k}} <\infty$ and $\sum a_nq_n$ does not converge because the general term of the series does not tend to $0$.

Kavi Rama Murthy
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