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- 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

$$

\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?
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