Run out of sums

Calculus Level 3

$\large 7 -\sqrt7 + \sqrt[3]{7} - \sqrt[4]{7} + \sqrt[5]{7} - \cdots$

The series $\displaystyle \sum_{j=1}^{\infty} a_j$ is said to be Cesàro summable , with Cesaro Sum $A$ , if the average value of its partial sums $\displaystyle s_k=\sum_{j=1}^k a_j$ tends to $A$ , meaning that $\displaystyle A=\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^ns_k$ .

Is the series above Cesàro summable?

No Yes

1 solution

Otto Bretscher
May 15, 2016

The sequence $s_{2n}$ of partial sums is increasing and bounded, and so is $\frac{1}{n}\sum_{k=1}^{n}s_{2k}$ , with $\lim_{n\to\infty}\left(\frac{1}{n}\sum_{k=1}^{n}s_{2k}\right)=L$ . Likewise, $\lim_{n\to\infty}\left(\frac{1}{n}\sum_{k=1}^{n}s_{2k-1}\right)=U$ . It follows that the Cesàro sum exists, with $A=\frac{L+U}{2}$ ; the answer is $\boxed{Yes}$

Yes sir i also did the same approach but this took me a lot of time

Abhishek Yadav - 5 years, 1 month ago

Comrade @Pi Han Goh and I had discussed similar problems before, so, I kept my solution brief.

Otto Bretscher - 5 years, 1 month ago

Run out of sums

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