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If $a_n$ is nondecreasing and converging to $L$, then $a_n\leq L$ for all $n$.

It is reversing the proof of the bounded sum test...

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    What do you mean exactly by "It is reversing the proof of the bounded sum test..."? Suggestions: Draw a picture, and think about the contrapositive. If there exists $k$ such that $a_k>L$, is there a neighborhood of $L$ that excludes the rest of the terms of the sequence?2012-07-15
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    This has nothing to do with [bounded-variation] or with partial fractions2012-07-15
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    To be completely honest I'm clueless. My professor just stated this and he just said to look at the bounded sum test and its proof...oh, I'm new so I just picked any tags, sorry.2012-07-15

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HINT Assume there exists a $m$ such that $a_m > L$, what can you then say about $\lim_{m \to \infty} a_m$?

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HINT: Suppose that some $a_n>L$. Then $$\lim_{k\to\infty}a_k=L

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Well, suppose there is an $n_0$ such that $a_{n_0} > L$. Now choose $n > n_0$ large enough that $$ L - (a_{n_0} - L) < a_n < L + (a_{n_0} - L) $$ and note that the second inequality violates the non-decreasing assumption.