For what values of ##\beta## does the series converge?

[Hall]

Homework Statement

$$

\sum_{k=1}^{\infty} k^{\beta} \left( \frac{1}{\sqrt k} - \frac{1}{\sqrt {k+1}}\right)

$$

Relevant Equations

Comparison Test.

For what values of β the following series converges
$$
\sum_{k=1}^{\infty} k^{\beta} \left( \frac{1}{\sqrt k} - \frac{1}{\sqrt {k+1}}\right)
$$

I thought of doing it like this
$$
\frac{k^{\beta} }{\sqrt k} - \frac{k^{\beta} }{\sqrt{k+1}}$$
$$0 \lt \frac{k^{\beta} }{\sqrt k} - \frac{k^{\beta} }{\sqrt{k+1}} \lt 2 \frac{k^{\beta}} { \sqrt{k} } ~~~~~~~~(1)$$
$$\textrm{Considering the series of the last term of the inequality above}$$
$$\sum_{k=1}^{\infty} 2 \frac{k^{\beta}} { \sqrt{k} } = \sum_{k=1}^{\infty} \frac{2}{ k^{1/2 - \beta} } $$
$$\textrm{For the above series to converge we must have}$$
$$\frac{1}{2} - \beta \gt 1$$
$$-\frac{1}{2} \gt \beta $$
$$\textrm{Since, for this beta the series of right most term of (1) converges, therefore, for this beta the following also converges}$$
$$\sum_{k=1}^{\infty} \frac{k^{\beta} }{\sqrt k} - \frac{k^{\beta} }{\sqrt{k-1}}$$


As my upper bound for $k^{\beta} \left( \frac{1}{\sqrt k} - \frac{1}{\sqrt {k+1}}\right)$ was very broad, therefore there can be some more values $\beta$ for which the series converges, but is it right that I have done?

 

[Orodruin]

First of all, you have a - instead of a + inside the root of the second term in your work. This makes your equation (1) wrong as with a - the given difference is smaller than 0. Now, assuming that you correct for that, yes, you can draw the conclusion that the series converges for those $\beta$, but you don't know if it converges for a larger range of $\beta$.

In particular, your bound on each term is way too broad - essentially requiring the sum over each of the two terms to converge separately. Instead, you should be working on combining the terms and the general behaviour of the combined term when you do.

 

[Hall]

Instead, you should be working on combining the terms and the general behaviour of the combined term when you do.

As you say Maussa,

$$
\begin{align*}

k^{\beta} \left(
\frac{1}{\sqrt k} - \frac{1}{\sqrt{k+1} } \right) \\
k^{\beta} \left(
\frac{\sqrt{k+1} - \sqrt{k}}{\sqrt k ~ \sqrt{k+1} } \right)\\
\textrm{Rationalizing the numerator} \\
k^{\beta} \frac{1}{ k \sqrt{k+1} + k \sqrt{k+1} }\\
k^{\beta} \frac{1}{ k \sqrt{k+1} + k \sqrt{k+1} } \approx \frac{k^{\beta}}{2 k^{3/2}} \\
\sum_{k=1}^{\infty} \frac{ k^{\beta}} { 2 k^{3/2} }~\textrm{converges if and only if}\\
3/2 - \beta \gt 1 \implies \beta \lt 1/2
\end{align*}
$$