1
$\begingroup$

How would I show that the following series converges or diverges.

$\sum_{j=1}^{\infty}\frac{\sqrt{j+1}-\sqrt{j}}{j+1}$

I am not sure what is the best test to show that it converges. Apparently it converges.

$\endgroup$
  • 2
    $\begingroup$ Start by rationalizing the numerator using $(x-y)(x+y)=x^2-y^2$ to get $\sqrt{j+1}-\sqrt{j} = \frac{1}{\sqrt{j+1}+\sqrt{j}}$. $\endgroup$ – Winther Oct 11 '15 at 17:54
  • 1
    $\begingroup$ @Winther I think that would almost do as an answer. $\endgroup$ – mickep Oct 11 '15 at 17:58
1
$\begingroup$

You may write $$ \sum_{j=1}^{\infty}\frac{\sqrt{j+1}-\sqrt{j}}{j+1}=\sum_{j=1}^{\infty}\frac1{(j+1){(\sqrt{j+1}+\sqrt{j}})} $$ and observe that, as $j \to \infty$, $$ \frac1{(j+1){(\sqrt{j+1}+\sqrt{j}})} \sim \frac1{2j^{3/2}} $$ leading to the convergence of the initial series.

$\endgroup$
2
$\begingroup$

$$ \frac{1}{\sqrt{j}}-\frac{1}{\sqrt{j+1}} = \frac{\sqrt{j+1}-\sqrt{j}}{\sqrt{j^2+j}}\geq\frac{\sqrt{j+1}-\sqrt{j}}{j+1}\geq 0 \tag{1}$$ hence: $$ 0 \leq \sum_{j\geq 1}\frac{\sqrt{j+1}-\sqrt{j}}{j+1}\leq\sum_{j\geq 1}\left(\frac{1}{\sqrt{j}}-\frac{1}{\sqrt{j+1}}\right)=\color{red}{1}.\tag{2}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.