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$$b_n=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}$$ $$n\geq1$$ $$c_n=\frac{1}{\sqrt1}+\frac{1}{\sqrt2}+...+\frac{1}{\sqrt{n}}$$ $$n\geq1$$

I have these 2 exercises and I need to prove the convergence of them by using the Cauchy series criterion :$$\lvert x_{n+p}-x_n \rvert \lt ε$$ Do i replace the terms $x_{n+p}$ and $x_n$ with: $$\lvert \frac{1}{(n+p)^2}-\frac{1}{n^2} \rvert \lt ε$$ $$\lvert \frac{1}{\sqrt(n+p)}-\frac{1}{\sqrt(n)} \rvert \lt ε$$? I used these 2 steps for both the series but I'm not sure if they're correct: $$b_{n+p}=\frac{1}{(n+1)^2}+\frac{1}{(n+2)^2}+...+\frac{1}{(n+p)^2}$$ $$c_{n+p}=\frac{1}{\sqrt(n+1)}+\frac{1}{\sqrt(n+2)}+...+\frac{1}{\sqrt(n+p)}$$ I used the criterion in that form but I'm not certain if this is correct or not.

wolly
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1 Answers1

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$b_{n+p}$ does not equal $\frac1{(n+p)^2}$ nor does

$$b_{n+p}\ne\frac1{(n+1)^2}+\dots+\frac1{(n+p)^2}$$

It equals

$$b_{n+p}=\frac1{1^2}+\dots+\frac1{(n+p)^2}$$

Likewise for $c_{n+p}$.

Now note that we have

$$|b_{n+p}-b_n|=\frac1{(n+1)^2}+\dots+\frac1{(n+p)^2}$$

What you want to show is that regardless of $p$, there exists an $N$ such that as long as $n>N$, then $|b_{n+p}-b_n|<\epsilon$. This can be shown with a little bit of creative telescoping. Note that we have:

$$\frac1{k^2}<\frac1{k(k-1)}=\frac1{k-1}-\frac1k$$

and hence

$$|b_{n+p}-b_n|<\frac1n-\frac1{n+p}<\frac1n<\frac1N<\epsilon$$

And hence $(b_n)_{n=1}^\infty$ is Cauchy.

Similarly, we have

\begin{align}|c_{n+p}-c_n|&=\frac1{\sqrt{n+1}}+\dots+\frac1{\sqrt{n+p}}\\&\ge\frac1{\sqrt{n+p}}+\dots+\frac1{\sqrt{n+p}}\\&=\frac p{\sqrt{n+p}}\\&\ge\frac p{\sqrt{p+p}}\tag{*}\\&=\sqrt{\frac p2}\end{align}

which is unbounded in $p$, where $(*)$ occurs when $p>n$. Thus, for any $\epsilon$ and $N$, then for all $n>N$ there is a $p>n$ where $|c_{n+p}-c_n|>\epsilon$. Thus $(c_n)_{n=1}^\infty$ is not Cauchy.

Simply Beautiful Art
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