Let $$a_n = \sum_{k=0}^n b_k$$ converge, then for some $N$ sufficiently large:

$$|b_n| \le \frac{1+\max\{|b_1|, |b_2|,...|b_N|\}}{2}$$

If the series converges that means $b_n$ converges to $0$. There exists an $N$ such that for all $n>N$: $|b_n| < 1/2$. Choose $\max\{|b_1|,\ldots, |b_n|\}$ then $|b_n|\le{1/2+\max(|b_1|, |b_2|,...|b_N|)}$ but I don't know how to divide them by $2$.


1 Answer 1


For $a_n$ to converge, the $b_n$ must converge to zero. This means that there is some $N$ such that $|b_n| \lt \frac 12$ for all $n \ge N$. The right side of your inequality is greater than $\frac 12$

  • $\begingroup$ Oh stupid me I already proved it. I didn't notice that N sufficiently large; I thought it was just bounded by the RHS for all N. $\endgroup$
    – Rainroad
    May 5, 2016 at 23:01

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