How to check the convergence or divergence of this alternating series: $\sum_{n=1}^{\infty}\frac{(-1)^n}{1+\sqrt{n}}$?

Question

Check the convergence or divergence of this alternating series: $$\sum_{n=1}^{\infty}\frac{(-1)^n}{1+\sqrt{n}}$$ My attempt:

I know that

$$\frac{1}{\sqrt n}>\frac{1}{n}\tag 1$$ we conclude that $\sum_{n=1}^{\infty}\frac{1}{\sqrt n}$ is divergent because harmonic series: $\sum_{n=1}^{\infty}\frac{1}{n}$ divergent

$$\frac{1}{\sqrt n}>\frac{1}{1+\sqrt n}\tag2$$ The second inequality doesn't implies that $\sum_{n=1}^{\infty}\frac{1}{1+\sqrt{n}}$ is divergent.

I am totally stuck here how to check the nature of given series. Any help would be greatly appreciated. Thanks

Answer 1

Let: $$a_n=\frac{1}{1+\sqrt{n}}$$ We have that $a_n$ is decreasing (you can easily check it with derivates) and: $$\lim_{n\to +\infty}a_n=0$$ Also, $a_n$ is always non-negative.

So, we can apply Leibniz's criteria for alternating sign series. Thus, $\sum_{n=0}^{+\infty}\frac{1}{1+\sqrt{n}}$ converges.

Answer 2

You can use the Alternating Series Test.

You only have to verify that $$ \frac{1}{1+\sqrt{n}} $$ decreases monotonically to limit of $0$. In other words, you have to justify that $$ \frac{1}{1+\sqrt{n+1}} < \frac{1}{1+\sqrt{n}} $$ for all $n$, and that $$ \lim_{n \to \infty} \frac{1}{1+\sqrt{n}} = 0. $$ Can you do that?

Answer 3

Notice that

$$\lim_{n\to\infty}\frac{1}{1+\sqrt{n}}=0,$$

and so by the alternating series test the series converges.

Answer 4

To make a sense $$\sum_{n=1}^{\infty}\frac{(-1)^n}{1+\sqrt{n}}=\\\frac{-1}{2}+\sum_{n=2}^{\infty}\frac{(-1)^n}{1+\sqrt{n}}\\= \frac{-1}{2}+\sum_{n=2k,k=1}^{\infty}(\frac{1}{1+\sqrt{n}}-\frac{1}{1+\sqrt{n+1}})\\= \frac{-1}{2}+\sum_{n=2k,k=1}^{\infty}(\frac{\sqrt{n+1}-\sqrt{n}}{1+\sqrt{n}+\sqrt{n+1}+\sqrt{n^2+n}})\\=$$ now notice that $\sqrt{n+1}-\sqrt{n}\leq \frac1{2\sqrt n}$ so

$$0 \leq \sum_{n=2k,k=1}^{\infty}(\frac{\sqrt{n+1}-\sqrt{n}}{1+\sqrt{n}+\sqrt{n+1}+\sqrt{n^2+n}})\leq \\\sum_{n=2k,k=1}^{\infty}(\frac{\frac1{2\sqrt n}}{1+\sqrt{n}+\sqrt{n+1}+\sqrt{n^2+n}})\leq \\\sum_{n=2k,k=1}^{\infty}(\frac{\frac1{2\sqrt n}}{\sqrt{n^2+n}})\leq \\\sum_{n=2k,k=1}^{\infty}(\frac{\frac1{2\sqrt n}}{\sqrt{n^2}}) $$ and it converges

Answer 5

The obvious way is to use the Leibniz's alternating series criteria. If you don't have the theorem at your disposition, you may do as follows:

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

and we have

\begin{align} 0 < \frac{1}{1+\sqrt{2k-1}}-\frac{1}{1+\sqrt{2k}} &= \frac{\sqrt{2k}-\sqrt{2k-1}}{(1+\sqrt{2k-1})(1+\sqrt{2k})}\\ & = \frac{1}{(1+\sqrt{2k-1})(1+\sqrt{2k})(\sqrt{2k}+\sqrt{2k-1})}\\ & \le \frac{1}{\sqrt{k}\sqrt{k}(\sqrt{k}+\sqrt{k})} = \frac{1}{2 k^{3/2}} \end{align}

and hopefully you already know that the series $\sum\limits_{k=1}^\infty \frac{1}{k^{3/2}}$ converges.

Answer 6

We investigate whether the series converges absolutely, so consider the series

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

We can see that this is very close to the series

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

which we will assume we know diverges (this is just the $p$-series with $p=\frac{1}{2}$, and it can be shown to diverge using, for example, the integral test, but that is beyond this answer). Now the limit comparison test tells us that

$$\lim_{n\to\infty}\frac{\frac{1}{1+\sqrt{n}}}{\frac{1}{\sqrt{n}}}=\lim_{n\to\infty}\frac{\sqrt{n}}{1+\sqrt{n}}=\lim_{n\to\infty}\frac{1}{\frac{1}{\sqrt{n}}+1}=1,$$

and so the series $S$ diverges. Now to check if the series in the question converges, consider

$$\lim_{n\to\infty}\frac{1}{1+\sqrt{n}}=0,$$

and so, as the sequence $\left\{\frac{1}{1+\sqrt{n}}\right\}_{n\in\mathbb{Z}^+}$ in the series clearly decreases monotonically (check this!), it follows from the alternating series test that the series

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

converges, and as it did not converge absolutely, it follows that it converges conditionally.