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Define the sum, $S_n = \sum\limits_{k=1}^{n} a_k$

Given that the sequence $a_k$ is bounded above, so can I show that $S_n$ is bounded above this way? :

Since $a_k$ is bounded above, there exists a real number $M$ such that $a_k \leq M$, for all $k$. So, we have, $$S_n = \sum_{k=1}^{n} a_k = a_1+a_2+a_3+...+a_n \leq M+M+M+\cdots+M =nM$$

Hence $S_n$ is also bounded above.

Is it correct?

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    $\begingroup$ Do you have to do all that work? Doesn't $S_n\le S_n$ prove that $S_n$ is bounded above? $\endgroup$ Commented Jun 15, 2021 at 2:42

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I choose $a_k=1$ for every $k$; you did not forbid it. Then we can have $M=1$. So $S_n=n$, which is not bounded above. Your proposition is not true.

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