2
$\begingroup$

I am trying to understand how this sum was transformed from

$$\sum_{n=1}^\infty \frac {\sqrt{n}}{n(n+1)}$$

to

$$ 1 + \sum_{n=2}^\infty \frac{\sqrt{n}-\sqrt{n-1}}{n} $$

I see that the index was changed from $n=1$ to $n=2$, thus requiring that the case for $ n=1$ be added but I get $\frac{1}{2}$. Not sure where the $1$ comes from and how they transformed the rest of the sum.

$\endgroup$
2
$\begingroup$

Note that

$$\begin{align} \sum_{n=1}^N\left(\frac{\sqrt n}{n(n+1)}\right)&=\sum_{n=1}^N\left(\frac{\sqrt{n}}n-\frac{\sqrt n}{n+1}\right)\\\\ &=\color{blue}{\sum_{n=1}^N\left(\frac{\sqrt{n}}n\right)}-\color{red}{\sum_{n=1}^N\left(\frac{\sqrt{n}}{n+1}\right)}\\\\ &=\color{blue}{1+\sum_{n=2}^N\left(\frac{\sqrt n}{n}\right)}-\color{red}{\sum_{n=1}^N\left(\frac{\sqrt{n}}{n+1}\right)}\\\\ &=\color{blue}{1+\sum_{n=2}^N\left(\frac{\sqrt n}{n}\right)}-\color{red}{\sum_{n=2}^N\left(\frac{\sqrt{n-1}}{n}\right)-\frac{\sqrt {N}}{N+1}}\\\\ &=1+\sum_{n=2}^N\left(\frac{\sqrt n-\sqrt{n-1}}{n}\right)-\frac{\sqrt{N}}{N+1}\\\\ \end{align}$$

Taking the limit as $N\to \infty$ shows that

$$\sum_{n=1}^\infty\left(\frac{\sqrt n}{n(n+1)}\right)=1+\sum_{n=2}^\infty\left(\frac{\sqrt n-\sqrt{n-1}}{n}\right)$$

as was to be shown!

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I have just edited the denominator on the transformed sum to reflect the equality at the moment it was transformed in the solution i'm referencing. This should make them equal. $\endgroup$ – john fowles Feb 16 '17 at 5:33
  • $\begingroup$ @johnfowles I've edited the answer accordingly. $\endgroup$ – Mark Viola Feb 16 '17 at 5:42
  • $\begingroup$ I have asked the question because I fail to understand the steps. I still don't understand why a 1 was pulled out and not the $\frac{1}{2}$. From my understanding of changing the lower limit is that if we change from $n=1$ to $n=2$ then we account for the $n=1$ case by adding it in. And why the change from $\infty$ to N? what's the significance of N? Is it substituted for $\infty$ in order to take the limit at $\infty$ instead. $\endgroup$ – john fowles Feb 16 '17 at 5:50
  • $\begingroup$ And I appreciate the solution ( which I already had) but I was hoping to get some insight into the solution as I fail to understand the reasoning behind the steps $\endgroup$ – john fowles Feb 16 '17 at 5:52
  • $\begingroup$ John, we work with the partial sums before passing to the limit. This avoids issues with convergence (note that the separate partial sums diverge, but the combined partial sums converge). I've edited to add a few more steps to facilitate. $\endgroup$ – Mark Viola Feb 16 '17 at 6:08

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.