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Let $a_n$ be a sequence in $\mathbb{R}$, $\phi: \mathbb{N} \rightarrow \mathbb{N}$ a monotonically increasing function with $\phi (0)=0$. Show that

$\sum _{n=0}^{\infty } a_n$ converges $\Rightarrow \sum _{i=0}^{\infty } {(\sum _{j=\phi (i)}^{\phi (i+1)-1 }a_{j})}$ converges.

My notes: Can you use that ${(\sum _{j=\phi (i)}^{\phi (i+1)-1 }a_{j})}$ has to be a cauchy sequence and can be as small as you want? Certainly it will be smaller than $a_n$ for all $n$ bigger than some $N_e$

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up vote 1 down vote accepted

I suppose you meant $\phi$ is strictly increasing, otherwise the inner sum could be an empty sum.

Let $b_i=\sum_{j=\phi(i)}^{\phi(i+1)-1} a_j$. For any $\epsilon >0$, you can find $N$ such that $$|a_j+\cdots +a_{j+k}|<\epsilon$$ for all $j\ge N, k>0$. Let $I$ be such that $\phi(I)\ge N$. Then for all $i\ge I$ you have $$ |b_i+\cdots +b_{i+k}|<\epsilon $$ for all $k>0$. This proves that your series is Cauchy and hence converges.

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When you take a fixed N you get

$\sum _{i=0} ^{N} \sum _{j=\phi(i)} ^{\phi(i+1)-1} a_j = \sum_{j=0}^{j=\phi(N+1)-1} a_j$

so basically the new sequence (meaning the partial summations) is a subsequence of the original one.

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If that step u wrote is true, I agree that the task is solved. However, it appears to me that the right sum is smaller. How can you change the index? – Listing Dec 2 '10 at 17:48
Think of the original sequence, when you add parentheses in some places. The function $\phi$ just tell you where you added them. The left sum is just adding first the elements in the parentheses and then add it to the total sum, while the right sum "forgets" the parentheses . – Prometheus Dec 2 '10 at 17:54

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