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$$\sum\frac{(-1)^n}{\sqrt{n+1}} \text{and} \sum\frac{1}{\sqrt{n+1}}$$

The first one is an alternating series, so it would just be: $$\sum (-1)^n\frac{1}{\sqrt{n+1}}\Rightarrow \;^\lim_{n\rightarrow\infty}\frac{1}{\sqrt{n+1}} \Rightarrow \frac{\frac1n}{\sqrt{1+\frac1n}}\Rightarrow \frac{0}{1}=0 = \text{convergent}$$

But for the second one I'm confused a little: $$^\lim_{n\rightarrow\infty}\frac{1}{\sqrt{n+1}} \Rightarrow \frac{\frac1n}{\sqrt{1+\frac1n}}\Rightarrow \frac{0}{1}=0 = \text{convergent}$$ or $$\sum\frac{1}{\sqrt{n+1}} \lt \frac{1}{\sqrt{n}}\Rightarrow\frac{1}{n^{1/2}} = \text{p-series}\frac12\lt1 = \text{divergent}$$

Is the second series convergent or divergent, and by what test(s)?

Note: in my last example I am comparing the series to a greater/divergent series

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  • $\begingroup$ yes, you are correct. $\endgroup$ – Jared Nov 23 '14 at 2:25
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First one is fine! but For $\displaystyle\sum_{n=0}^{\infty}\frac{1}{\sqrt{n+1}}$ You've to compare it with smaller but divergent series to show divergence

for $n\gt2$ have $$\frac{1}{n}\lt\frac{1}{\sqrt{n+1}}$$ and $$\sum_{n=1}^{\infty}\frac1n\le\sum_{n=0}^{\infty}\frac{1}{\sqrt{n+1}}$$ The first series in left side is Harmonic series which is obviously divergent!

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  • $\begingroup$ Integrator! :P, why not simply use $\frac{1}{n+1}\le\frac{1}{\sqrt{n+1}}$? $\endgroup$ – Swapnil Tripathi Nov 23 '14 at 2:29
  • $\begingroup$ Exactly. You wouldn't have to use "for $n>2$" :P $\endgroup$ – Swapnil Tripathi Nov 23 '14 at 2:32
  • $\begingroup$ It was already correct, mate! :) $\endgroup$ – Swapnil Tripathi Nov 23 '14 at 2:39
  • $\begingroup$ so the answer would be divergent for n>2? $\endgroup$ – Jessica Nov 23 '14 at 3:12
  • $\begingroup$ yep, nevermind. Thanks! $\endgroup$ – Jessica Nov 23 '14 at 3:22
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You are making a big mistake. The fact that $$\lim\limits_{n\rightarrow \infty}a_n = 0 $$ does not imply that $\sum a_n$ is convergent. On the other hand, if the limit is nonzero, we can conclude that the sum is divergent.

The first series is an alternating series and converges by the Leibniz alternating series test. For the second series compare $$\frac{1}{\sqrt{n+1}}\geq \frac 1n $$ when $n\geq 2$ and $\sum\frac1n$ is the classic example of a divergent series, therefore so is the second series.

ADDED: Since you're confused about the p-test. $$\sum\limits_{n=1}^\infty \frac{1}{\sqrt{n+1}} = \sum\limits_{n=2}^\infty \frac{1}{\sqrt{n}} = \sum\limits_{n=2}^\infty \frac{1}{n^{1/2}}$$ hence the series diverges by the p-test.

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  • $\begingroup$ does my use of the p-series work to prove it's divergence? $\endgroup$ – Jessica Nov 23 '14 at 2:28
  • $\begingroup$ Yes, it looks fine to me. $\endgroup$ – Eff Nov 23 '14 at 2:30
  • $\begingroup$ @Jessica: You in your proof used $\frac{1}{\sqrt{n+1}}> \frac{1}{\sqrt{n}}$. Do you think it is correct? $\endgroup$ – Swapnil Tripathi Nov 23 '14 at 2:35
  • $\begingroup$ What is $a_n$ and $b_n$ respectively? If $a_n\geq b_n$ and $\sum b_n$ diverges, then so does $\sum a_n$. If $a_n\leq b_n$ and $\sum b_n$ converges, then so does $\sum a_n$ (assuming that $a_n,b_n\geq 0$). $\endgroup$ – Eff Nov 23 '14 at 2:37
  • $\begingroup$ $a_n = \frac{1}{\sqrt{n+1}}$ and $b_n = \frac{1}{\sqrt{n}}$ $\endgroup$ – Jessica Nov 23 '14 at 2:39
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Hint: In the second series

$\frac{1}{\sqrt{n+1}}\ge\frac{1}{n+1}$

First converges by alternating series test : https://en.wikipedia.org/wiki/Alternating_series_test

and uses just what you've shown.

IMPORTANT: It seems that you think that $\sum a_n$ converges $\iff$ $\lim_{n\to\infty}a_n= 0$.

It is not true that $\lim_{n\to\infty}a_n= 0\Rightarrow\sum a_n$ converges. Consider $\sum\frac{1}{n}$

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Your confusion is that the second sequence converges to 0:

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

For the series to converge, the sequence must converge to $0$ (so that you are eventually adding $0$), but it's not sufficient (e.g. p-series). You correctly compared your series with a divergent series that was less than your series, so this is a correct application of the comparison test.

Edit:

Jessica, yes you are correct, it was a greater series (which means the comparison is not correct). Although in this case it's a fairly trivial thing, you can simply use a change of indexes:

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

I think the more interesting thing would be prove that:

$$ \sum \frac{1}{(x^d + ...)^r} $$

is convergent when $dr > 1$ (and $d$ is the largest--most positive--exponent of the polynomial) and divergent otherwise (in your case you would have $1\cdot\frac{1}{2} = \frac{1}{2} \leq 1$ and therefore divergent).

The key here is to compare with:

$$ \sum \frac{1}{(x^d)^r} = \sum \frac{1}{x^{dr}} $$

For simplicity of my argument, lets assume that $d, r > 0$, this means we have:

$$ \frac{1}{(x^d + ...)^r} = \frac{1}{\left(x^d\left(1 + \frac{...}{x^d}\right)\right)^r} = \frac{1}{x^{dr}\left(1 + \frac{...}{x^d}\right)^r} $$

The reason its important to assume $d > 0$ (and is the greatest, most positive, exponent) is that $\frac{...}{x^d}$ tends towards zero. This means that there exists some value, $l$ for $x = x_{max}$ such that $l > \frac{...}{\left(x_{max}\right)^d}$, which means at some point we can say that:

$$ \frac{1}{x_{max}^{dr}}\frac{1}{\left(1 + \frac{...}{\left(x_{max}\right)^d}\right)^r} > \frac{1}{\left(x_{max}\right)^{dr}}\frac{1}{\left(1 + l\right)^r} $$

Since $\frac{1}{(1 + l)^r}$ is some constant, this amounts to proving the p-series case.

For your particular case this would be:

$$ \frac{1}{(n + 1)^{\frac{1}{2}}} = \frac{1}{n^{\frac{1}{2}}\left(1 + \frac{1}{n}\right)^{\frac{1}{2}}} $$

We can choose $x_{max} = 2$ and $l = 1$. Thus giving:

$$ \frac{1}{\left(1 + \frac{1}{n}\right)^\frac{1}{2}} \stackrel{?}{>} \frac{1}{(1 + 1)^\frac{1}{2}} \longrightarrow \frac{1}{\sqrt{n + 1}} \stackrel{?}{>} \frac{1}{\sqrt{2n}} $$

This inequality is true because $1 > \frac{1}{n}$ for all $n > 1$ and therefore you are dividing by a larger number on the right hand side, making the right hand side smaller. Since the right hand side, which is smaller diverges, so too does the larger, left hand side.

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    $\begingroup$ I beleive I actually compared it with a divergent series that was greater than my series. $\endgroup$ – Jessica Nov 23 '14 at 2:54
  • $\begingroup$ @Integrator what would be a good comparison? $\endgroup$ – Jessica Nov 23 '14 at 3:09
  • $\begingroup$ @Jessica see my edit, you were correct, your comparison is not quite correct, but it is "basically" correct. $\endgroup$ – Jared Nov 23 '14 at 4:30

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