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I need to prove or disprove the statement. I think the statement is true.

My attempt at a proof:

From the definition of convergence:

$$\forall \epsilon > 0 \quad \exists N \in \mathbb{N} \quad st \quad n,m\geq N \implies \left|\frac{a_{m+1}}{1+a_{m+1}}+\dots+\frac{a_{n}}{1+a_{n}}\right|<\epsilon$$

Since there is a finite amount of denominators and $a_n>0$ we can take the maximum:

$$g = \max\{1+a_{m+1}, \dots, 1+a_n\}$$

$$\frac{1}{g}\left|a_{m+1},\dots,a_n\right|\leq\left|\frac{a_{m+1}}{1+a_{m+1}}+\dots+\frac{a_{n}}{1+a_{n}}\right|<\epsilon$$

and that completes the proof since $\epsilon$ is arbitrary. Does this make sense?

Margaret S.
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  • Looks correct, although you might want to change “definition of convergence” to “definition of Cauchy sequence” since, while the two coincide for sequences in $\mathbb{R}$, the equivalence does not hold in other metric spaces. – user1892304 May 01 '16 at 21:38
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    If $x \in [0,1]$ then ${1 \over 2} x \le {x \over 1+x } \le x$. – copper.hat May 01 '16 at 21:40

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