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We have to determin the convergence of the following sum: \begin{equation} \sum\limits_{n=1}^{\infty}\frac{-(n-1)}{n\sqrt{n+1}} \end{equation}

And we've seen multiple ways to determine if the sum converges or diverges, but they only work with non negative terms. So tests like d'Alemberts ratio test, Cauchy's root test or Raabe's test wont work. How would I go about determining the convergence of this sum?

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  • $\begingroup$ then get rid of the $-1$, since $\sum \frac{1}{n^{3/2}}$ converges. so you have $\sum_n \frac{1}{\sqrt{n}}$, which diverges $\endgroup$ – mathworker21 Jan 7 at 16:54
  • $\begingroup$ @CalvinGodfrey I don't think that's "without loss of generality" $\endgroup$ – mathworker21 Jan 7 at 16:55
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You can use Integral test:

$$f(n)=\dfrac{n-1}{n\sqrt{n+1}}$$

When you plot a graph, you will notice that $f(n)$ is positive, continuous and decreasing from $n=4$

So, $$\int_4^\infty \dfrac{n-1}{n\sqrt{n+1}}=\mbox{diverges}$$

So, by integral test, the series diverges

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  • $\begingroup$ I have 2 questions about this: first, why did you take out the negative sign? Can you take out the negative sign and when you do this and the remaining part converges or diverges, does it still converge or diverges, and second why can you start at 4 and not at 1 in your integral? $\endgroup$ – Viktor Jan 7 at 16:59
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    $\begingroup$ @Viktor first, I took the constant out, here $(-1)$, and then performed the test. This doesn't change the result. I took the boundary from $4$ only after checking if $f(n)$ is positive, continuous and decreasing. $\endgroup$ – Key Flex Jan 7 at 17:03
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    $\begingroup$ Okay thanks i think i understand now! $\endgroup$ – Viktor Jan 7 at 17:03
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For $n>2$, $$\frac{n-1}{n\sqrt{n+1}}>\frac1{2\sqrt{n+1}} $$ As $\sum \frac1{2\sqrt{n+1}}$ diverges. so does $\sum\frac{n-1}{n\sqrt{n+1}}$ and also $\sum\frac{-(n-1)}{n\sqrt{n+1}}$.

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You may also use the comparison test.

Intuitively, as $n$ gets larger, $n-1$ is essentially $n$, and $\sqrt{n+1}$ is essentially $\sqrt{n}$. That means you can compare $\frac{n-1}{n\sqrt{n+1}}$ with $\frac{n}{n\sqrt{n}}=\frac{1}{\sqrt{n}}$

Formally, you can find a constant $C>0$ such that: $$\frac{n-1}{n\sqrt{n+1}}>C\frac{1}{\sqrt{n}}, \text{ for }n\geq 2$$

(what could that constant be?)

Now since $\sum_{n=1}^\infty\frac{1}{\sqrt{n}}$ diverges, that means $\sum_{n=1}^\infty\frac{n-1}{n\sqrt{n+1}}$ diverges.

Obs.: For any non-zero constant $\lambda$, we have that $\sum_{n=1}^\infty \lambda\, a_n$ converges $\Leftrightarrow \sum_{n=1}^\infty a_n$ converges (this should be a simple exercise). In particular, the negative sign in $\frac{-(n-1)}{n\sqrt{n+1}}$ is not relevant to convergence).

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