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I'm trying to solve the below exercise.

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Could you please verify whether my proof is correct or not? Thank you so much!

My attempt:

It is easy to verify that $\sum x_{k}<\infty \implies \sum x_k/(x_k+1)<\infty$. Next we prove that $\sum x_k/(x_k+1)<\infty \implies \sum x_{k}<\infty$. To simplify notation, I write $\lim x_k$ for $\lim_{k \to \infty} x_k$.

Clearly, $\lim x_k = 0$. If not, $\lim x_k = a \neq 0$. Then $\lim x_k/(x_k+1) = a/(a+1) \neq 0$. This contradicts the fact that $\sum x_k/(x_k+1)<\infty$. As a result, for all $\epsilon_1 >0$ there exists $N_1 \in \mathbb N$ such that $x_k < \epsilon_1$ for all $k \ge N_1$. As such, $x_k \le (\epsilon_1+1 )x_k/(x_k+1)$ for all $k \ge N_1$.

By Cauchy's convergence test: given $\epsilon >0$, there exists $N_2 \in \mathbb N$ such that $\sum_{k=m}^n x_k/(x_k+1) < \epsilon/(\epsilon_1+1)$ for all $n>m \ge N_2$. Take $N=\max\{N_1,N_2\}$, we have $$\begin{cases}x_k \le (\epsilon_1+1 )x_k/(x_k+1) \\ \sum_{k=m}^n x_k/(x_k+1) < \epsilon/(\epsilon_1+1) \end{cases} \quad k \ge N, n>m \ge N$$

As such, we have for all $n>m \ge N$: $$\begin{aligned}\sum_{k=m}^n x_k &\le \sum_{k=m}^n \dfrac{(\epsilon_1+1)x_k}{x_k+1} &&= (\epsilon_1+1)\sum_{k=m}^n \dfrac{x_k}{x_k+1}\\ &\le (\epsilon_1+1) \epsilon/(\epsilon_1+1) &&=\epsilon\end{aligned}$$

By Cauchy's convergence test, $\sum x_{k}<\infty$.

Update: On the basis of suggestion from comments, I fixed the proof that $\lim x_k = 0$ here.

  • We have $\lim x_k \neq \infty$. If not, $\lim x_k/(1+x_k) = 1$. Then $\lim_{k \to \infty} x_k/(x_k+1) \neq 0$ which is a contradiction.

  • We have $\lim x_k$ does exists. If not, there are two subsequences $(x_{k_n})_{n \in \mathbb N}$ and $(x_{k_m})_{m \in \mathbb N}$ such that $\lim_{n \to \infty} x_{k_n} =:a \neq b:= \lim_{m \to \infty} x_{k_m}$. Then $$\lim_{n \to \infty} x_{k_n}/(1+x_{k_n}) = a/(a+1) \neq b/(b+1) \lim_{m \to \infty} x_{k_m}/(1+x_{k_m})$$

Then $\lim_{k \to \infty} x_k/(x_k+1)$ does not exist, which is a contradiction.

  • We have $\lim x_k = 0$. If not, $\lim x_k = a \neq 0$. Then $\lim_{k \to \infty} x_k/(x_k+1) = a/(a+1) \neq 0$, which is a contradiction.
Akira
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    Simpler: Eventually (say for $k \geq N$) we have $x_k<1$. Then $x_k/(1+x_k) \geq x_k/(1+1) = x_k/2$. Thus $\sum_{k=N}^\infty x_k/2 $ converges by the comparison test, so it follows that $\sum_{k=1}^\infty x_k$ converges. – kccu Aug 27 '19 at 19:41
  • Is my attempt fine @kccu? – Akira Aug 27 '19 at 19:44
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    Yes, I don't see any mistakes. – kccu Aug 27 '19 at 19:47
  • Thank you so much @kccu for your support ^^ – Akira Aug 27 '19 at 19:48
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    The opposite of $\lim x_k=0$ is not $\lim x_k=a\ne 0$. – Hagen von Eitzen Aug 27 '19 at 19:55
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    Perhaps easier: If $y_k=\frac{x_k}{x_k+1}$, then $x_k=\frac1{\frac1{y_k}-1}=\frac{y_k}{1-y_k}$. If $\sum y_k$ converges, then $y_k\to 0$ and for almost all $k$, $y_k<\frac12$, so $x_k<2y_k$. – Hagen von Eitzen Aug 27 '19 at 19:58
  • @Akira Hagen von Eitzen has a good point. Your proof that $x_k\to 0$ is not valid. It is true that $x_k \to 0$, but you will need to prove it a different way. – kccu Aug 27 '19 at 20:06
  • Hi @kccu, I have fixed that part. Please have a check on it! – Akira Aug 27 '19 at 20:20
  • Note that this has been solved here https://math.stackexchange.com/q/2862826/42969 and here https://math.stackexchange.com/q/2598643/42969. – Martin R Aug 28 '19 at 07:20

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