# Is the “reverse” sum summable?

Suppose we have a summable series $\sum_{n=0}^{\infty}a_n$, where $a_n$ are positive. Define new $b_n=\sum_{k=0}^{\infty} a_k-\sum_{k=0}^na_k$. Does $\sum_{n=0}^{\infty} b_n$ exist?

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No, it need not exist. If it does exist, it has to equal $\sum_n na_n$, since each term $a_n$ is included $n$ times in the sum. Thus, for example $a_n=1/(n+1)^2$ is summable, but $b_n$ would have to sum to the sum of $n/(n+1)^2$, which doesn't converge.