Cauchy’s first theorem on limits: If a sequence $\{x_n\}$ converges to l, then the sequence $\{y_n\}$ also converges to l.
Where, $y_n=\frac{x_1+x_2+\dots+x_n}{n}$
- $ \lim_{n\to\infty} \frac{1}{n}(1+\frac{1}{3}+\frac{1}{5}+\dots+\frac{1}{2n-1})=0$
Now in this example
$\{x_n\}=\frac{1}{2n-1}\implies \lim_{n\to\infty}\{x_n\}=\lim_{n\to\infty}\frac{1}{2n-1}=0 $
Hence by the Cauchy’s first theorem
$\implies \lim_{n\to\infty} \frac{1}{n}(1+\frac{1}{3}+\frac{1}{5}+\dots+\frac{1}{2n-1})=0$
- $ \lim_{n\to\infty} (\frac{1}{(n+1)^2}+\frac{1}{(n+2)^2}+\dots+\frac{1}{(2n)^2})=0$
$ \lim_{n\to\infty} (\frac{1}{n^2(1+\frac{1}{n})^2}+\frac{1}{n^2(1+\frac{2}{n})^2}+\dots+\frac{1}{n^2(2)^2})=0$
$ \lim_{n\to\infty} \frac{1}{n}(\frac{1}{n(1+\frac{1}{n})^2}+\frac{1}{n(1+\frac{2}{n})^2}+\dots+\frac{1}{4n})=0$
Now in this example
$\{x_n\}=\frac{1}{4n}$
$\implies \lim_{n\to\infty}\{x_n\}=\lim_{n\to\infty}\frac{1}{4n}=0 $
Hence by the Cauchy’s first theorem
$ \lim_{n\to\infty} (\frac{1}{(n+1)^2}+\frac{1}{(n+2)^2}+\dots+\frac{1}{(2n)^2})=0$
- $ \lim_{n\to\infty} (\frac{1}{n^2}+\frac{1}{(n+1)^2}+\frac{1}{(n+2)^2}+\dots+\frac{1}{(2n)^2})=0$
I think this example is similar as above example.
- $ \lim_{n\to\infty} (\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}}+\frac{1}{\sqrt{n+2}}+\dots+\frac{1}{\sqrt{2n}})=\infty$
Multiply and divide by n so we get,
$ \lim_{n\to\infty} \frac{1}{n}(\frac{n}{\sqrt{n}}+\frac{n}{\sqrt{n+1}}+\frac{n}{\sqrt{n+2}}+\dots+\frac{n}{\sqrt{2n}})$
Now $\{x_n\}=\frac{n}{\sqrt{2n}}=\frac{\sqrt{n}\sqrt{n}}{\sqrt{2n}}=\frac{\sqrt{n}}{\sqrt{2}}$
$\lim_{n\to\infty}\{x_n\}=\lim_{n\to\infty}\frac{\sqrt{n}}{\sqrt{2}}=\infty$
Hence by the Cauchy’s first theorem
$ \lim_{n\to\infty} (\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}}+\frac{1}{\sqrt{n+2}}+\dots+\frac{1}{\sqrt{2n}})=\infty$
Are the above right and if not could you give me a hint?
