If there exists $r \in \mathbb{R}$ with $r < 1, K \in \mathbb{N}$, such that $|X_n|^{1/n} < r$ for all $n > K$, then the series $X_n$ (Summation of $X_n$ from $n$ to $\infty$) is absolutely convergent.

We know that $|X_n| < r^n$ for all $n > K$ so sum of all terms would be bounded by sum of finite terms up to $k-1$ and sum of geometric series. Thus, this is bounded above and increasing $\implies$ convergent.

However, How can I show that this is absolutely convergent by using Cauchy Criteria?


If $\sum_{n=1}^{\infty}|y_n|<\infty$ and if $|x_n|\leq |y_n|$ for all but finitely many $n$ , then $\sum x_n$ converges absolutely. This is because of the Cauchy condition for absolute convergence: $ \lim_{n\to \infty} E(n)=0$ where $E(n)=\sup_{m \geq n} \sum_{j=n}^{j=m} |y_j|$ This implies that $\lim_{n\to \infty} \sup_{m\geq n}|\sum_{j=n}^{j=m}x_j|\leq \lim_{n\to \infty}\sup_{m\geq n}\sum_{j=n}^{j=m} |x_n| \leq \lim_{n\to \infty} E(n)=0 $ . And a generalization of your Q is the Hadamard radius formula: Let $s=\lim_{n\to \infty}\sup_{m>n}|A_m|^{1/m}.$ If $|x|<1/s$ (where we put $1/s=\infty$ when $s=0$) then $\sum_n A_n x^n$ converges absolutely.

  • $\begingroup$ What is the Yn?? $\endgroup$ – jessie Oct 14 '15 at 3:38
  • $\begingroup$ ($y_n)_n$ is any sequence which is absolutely summable. In your Q you can choose $ y_n=r^n$. $\endgroup$ – DanielWainfleet Oct 14 '15 at 4:46

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