Determining the convergence of $ \sum\limits_{n=1}^{\infty}\frac{-(n-1)}{n\sqrt{n+1}}$

Question

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?

Answer 1

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

Answer 2

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 $\sum\frac{-(n-1)}{n\sqrt{n+1}}$.

Answer 3

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).