Prove that if $\sum_{n = 1}^{\infty} a_{n}$ converges, then $\sum_{k = 1}^{\infty} A_{k}$ converges(2)

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Suppose that $\{a_n\}$ be a sequence of real numbers.let $\{n_{k}\}$ be an increasing sequence of positive integers, and let $A_{k} = a_{n_{k}} + a_{n_{k}+1}+ ... +a_{n_{k +1}-1}$, for each $k \in \mathbb{N}$. Prove that if $\sum_{n = 1}^{\infty} a_{n}$ converges, then $\sum_{k = 1}^{\infty} A_{k}$ converges. Give an example to show that the converse is not true.

Can anyone explain for me in the term $a_{n_{k+1} -1}$ in the definition of $A_k$ we take a decreasing index (i.e why we subtract 1 and not add 1 to $n_{k+1}$ as we continued adding 1 to $n_{k}$ ) or this is a typo?

Also can anyone give me a numeric example for the direction we want to prove ?

Also I received a hint about the proof that I should use Cauchy criterion, could anyone provide me with a little bit more details in this direction?

Those are related links:

1- Prove that if $\sum_{n = 1}^{\infty} a_{n}$ converges, then $\sum_{k = 1}^{\infty} A_{k}$ converges.

2-Prove that if $\sum_{n=1}^\infty a_n$ converges then $\sum_{k=1}^\infty A_k$ converges

Thank you!

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It is important not to include $a_{n_{k+1}}$ in $A_k$ to avoid this term getting repeated in $A_k$ as well as $A_{k+1}$. To prove the convergence of $\sum A_k$ note that we can write $A_k$ as $S_{n_{k+1}} -S_{n_k}$ where $\{S_n\}$ is the partial sum sequence of $\sum a_n$. Hence $A_1+A_2+\cdots+A_N= (S_{n_2} -S_{n_1})+(S_{n_{3}}- S_{n_2})+\cdots+S_{n_{N+1}} -S_{n_N}=S_{n_{N+1}}-S_{n_1}$. Since $\{S_n\}$ has a finite limit it is now clear that $\sum A_k$ converges. Example: take $n_k=2k$ as see what happens. To prove that the converse is not true take $n_k=2k$. Consider the series $1+(-1)+1+(-1)+\cdots$ . This series is not convergent but $A_1=0,A_2=0,\cdots$. Hence $\sum A_k$ is convergent even though $\sum a_k$ is not.