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Consider a non-negative sequence $t_1,t_2,...$ that is also bounded above? Suppose that the sequence is "pseudo non-increasing" in the sense that $t_{n+1} \leq t_n + e_n$ where $e_1 + e_2 + ...$ is finite.

Is the sequence necessarily convergent? I see that this is obviously true for the trivial case where $e_n=0$ for all $n$.

If yes, can this be proved under the weaker hypothesis that $e_n$ goes to $0$ as $n$ goes to $\infty$ (also I am not sure if one needs the upper bound). If not, can this be redeemed using some additional assumptions?

Ben Grossmann
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  • For your second question, consider the sequence $$ 0,1,\textstyle{1\over2},0,\textstyle{1\over3},\textstyle{2\over3}, 1,\textstyle{3\over4},\textstyle{2\over4},\textstyle{1\over4},0,\textstyle{1\over 5},\ldots $$ – David Mitra Aug 25 '14 at 14:31

1 Answers1

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Let $l = \limsup t_n$ and pick some $k_n$ so that $t_{k_n} \to l$.

Fix some $\epsilon >0$.

Then there exists some $N$ so that $t_{k_n} >l-\epsilon$ for all $n >N$. We can also assume that $t_{n} < l+\epsilon$ for all $n> k_N$, by the definition of lim sup.

Now, since $\sum e_i$ is convergent, it is Cauchy. Therefore, there exists some $M$ so that for all $n >m >M$ we have $$|e_{m+1}+e_{m+2}+..+e_n| <\epsilon \,.$$

Let $K =\max \{ k_N, M \}$.

Then, for all $n >K$ we have

Pick some $l$ so that $k_l > n$. Then

$$t_n=t_n-t_{k_l}+t_{k_l} \geq t_{k_l} -| t_n-t_{k_l} | \geq l-\epsilon -\epsilon=l-2 \epsilon \,.$$

Moreover, as $n>k_N$ we also have $$t_n < l+\epsilon \,.$$

This proves that for all $n>K$ we have $$|l-x_n| <2 \epsilon \,.$$

N. S.
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  • Thanks. I think you meant K=max(k_N,M) and that t_n < l + \varepsilon for sufficiently large n (second last line). –  Aug 26 '14 at 03:49