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Pleas, give a hint with this exercise.

If $(a^k)$ is a non increasing sequence with $a_k\to 0$. Is it true that

The series $\displaystyle\sum_{n=1}^{\infty}a_n$ is convergent iff the series $\displaystyle\sum_{k=0}^{\infty}2^{2^{k}}a_{2^{2^{k}}}$ is convergent? I am trying, to find an inequality,to use comparison and Cauchy condensation theorem, but i get nothing.

Thank you

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    $(a^k)$ or $(a_k)$? $2^{2^k}$ or $2^{k}$ ? – 2017-02-20
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    $(a_k)$ should be sequence of strictly positive numbers as well! – 2017-02-20
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    (1) Perhaps you should lower that index from $\;a^k\;$ to $\;a_k\;$ ...? (2) If the sequence is non-increasing and $\;a_k\to0\;$ then it must be a **positive** sequence so you can directly use the condensation test... – 2017-02-20
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    It seems like there's probably a typo in the general term's index: shouldn't it be $\;a_{2^n}\;$ instead of $\;a_{2^{2^n}}\;$ ? And please do address the comments . – 2017-02-20

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By the Cauchy condensation test $$ \sum_{n=1}^\infty a_n $$ converges if and only if $$ \sum_{k=1}^\infty 2^ka_{2^k} $$ converges. Applying the test again, this is equivalent to the convergence of $$ \sum_{j=1}^\infty 2^j\,2^{2^j}a_{2^{2^j}}=\sum_{j=1}^\infty 2^{2^j+j}a_{2^{2^j}} $$ Now use $0