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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

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6 Answers 6

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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?

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  • $\begingroup$ so can i say that it converges contionally? $\endgroup$
    – TShiong
    Feb 11, 2022 at 18:18
  • $\begingroup$ You can say it converges. If the absolute series (the series with absolute value of terms) converges, you can go further and say that it converges absolutely. Otherwise, you can conclude that converges conditionally. $\endgroup$ Feb 11, 2022 at 18:21
  • $\begingroup$ Because of the alternating nature of the terms, the partial sums bounce up and down in smaller and smaller increments, settling on the limit. But, if you consider the absolute version of the series, all of the "down bounces" are also up bounces, so it may not converge. $\endgroup$ Feb 11, 2022 at 18:22
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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.

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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

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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.

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Notice that

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

and so by the alternating series test the series converges.

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  • $\begingroup$ This is not sufficient, one needs to check all conditions. $\endgroup$ Feb 11, 2022 at 18:21
  • $\begingroup$ does it converge absolutely or conditionally? $\endgroup$
    – TShiong
    Feb 11, 2022 at 18:22
  • $\begingroup$ @LutzLehmann I think the other conditions are quite self-evident here, but you are right $\endgroup$
    – Lorago
    Feb 11, 2022 at 18:22
  • $\begingroup$ @ShivaVenkata Since you care about absolute convergence here, which I didn't know, I'll make another answer to adress that $\endgroup$
    – Lorago
    Feb 11, 2022 at 18:23
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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.

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