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I have $$M_{n}=\sup_{x\in\left[1,2\right]}\left|\frac{x}{\left(x+1\right)^{n}}\right|$$

How do I evaluate this? Am I supposed to fix $n\in\mathbb{N}$ before trying to find the supremum? I tried doing that, but I see that if $n=1$, then the supremum occurs at $x=2$, but if $n\neq1$, then the supremum occurs at $x=1$. Or, am I not supposed to fix $n$ at all?

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You are supposed to fix $n$ before taking the supremum over $x \in [1,2]$.

Yes, the supremum occurs at $x = 2$ if $n=1$, but at $x = 1$ if $n=2$. However, if you plot the graphs, you'll discover that the supremum actually occurs at $x = 1$ for all $n \geq 2$. (Presumably you can also prove this rigorously by standard techniques, e.g. by differentiating...)

Since your ultimate goal is to apply the Weierstrass M-test, it doesn't matter if you throw away a single term. So you may as well compute $M_n$ for $n \geq 2$ only, and then try to determine whether or not $\sum_{n=2}^\infty M_n$ is a convergent series.

Kenny Wong
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    What it also work if I did $\sum_{n=1}^{1}M_{n}+\sum_{n=2}^{\infty}M_{n}=\frac{2}{3}+\frac{1}{2}=\frac{7}{6}$? In the first sum, it is the supremum at $n=1$, which is when $x=2$, and in the second sum, it is the supremum at $n\ge 2$, which is at $x=1$. –  May 06 '17 at 21:07
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    @Jake yes, that is fine too. – Kenny Wong May 06 '17 at 21:08