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...
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...
HINT Assume there exists a $m$ such that $a_m > L$, what can you then say about $\lim_{m \to \infty} a_m$?
HINT: Suppose that some $a_n>L$. Then $\lim_{k\to\infty}a_k=L
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.