# Convergence of Sequences

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