Does the Alternating Series Test show convergence for this series?

[murshid_islam]

Homework Statement

To determine whether a series is convergent using the Alternating Series Test

Relevant Equations

$$\sum\limits_{n=1}^{\infty} (-1)^n\frac{\sqrt n}{2n+3}$$

The Alternating series test has to be used to determine whether this series converges or diverges: \sum\limits_{n=1}^{\infty} (-1)^n\frac{\sqrt n}{2n+3}

Here's what I have done:

Let a_n = \frac{\sqrt n}{2n+3}. Therefore, a_{n+1} = \frac{\sqrt {n+1}}{2n+5}

Now, for a_{n+1} to be less than or equal to a_n \\,
\frac{\sqrt {n+1}}{2n+5} \leq \frac{\sqrt n}{2n+3} \\
\Rightarrow \sqrt {\frac{n+1}{n}} \leq \frac{2n+5}{2n+3} < \frac{4n+6}{2n+3} \\
\Rightarrow \sqrt {\frac{n+1}{n}} < 2 \\
\Rightarrow \frac{n+1}{n} < 4 \\
\Rightarrow 3n > 1 \\
\Rightarrow n > \frac{1}{3}

Therefore, a_{n+1} \leq a_n for all n \geq 1

And \lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{\sqrt n}{2n+3} = \lim_{n \to \infty} \frac{1}{2 \sqrt n+ \frac{3}{\sqrt n}} = 0

Therefore, by Alternating Series Test, the series is convergent.

Is this ok?
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[fresh_42]

I know it as Leibniz criterion, but anyway. You have to write it backwards: from $n> \frac{1}{3}$ to $a_{n+1}<a_n$, since that it is what you want to show. You have shown an implication of an already convergent series, the wrong logical direction.

 

[Mark44]

You have shown an implication of an already convergent series, the wrong logical direction.--

I disagree. To show that an alternating series converges, all you need to do is to verify the two conditions:

1. $\lim_{n \to \infty} a_n = 0$
2. The terms are decreasing; i.e., $a_{n + 1} \le a_n$

In the OP's work on determining the latter condition, I believe that all the steps are reversible.

Another way to check that the terms are decreasing is to look at the limit of ##\frac{a_{n+1}}{a_n}. if this limit is less than 1, the sequence is decreasing.
The OP has done both of these, and has concluded that the series converges. I agree with his conclusion.

 

[member 587159]

I disagree. To show that an alternating series converges, all you need to do is to verify the two conditions:

1. $\lim_{n \to \infty} a_n = 0$
2. The terms are decreasing; i.e., $a_{n + 1} \le a_n$

In the OP's work on determining the latter condition, I believe that all the steps are reversible.

Another way to check that the terms are decreasing is to look at the limit of ##\frac{a_{n+1}}{a_n}. if this limit is less than 1, the sequence is decreasing.
The OP has done both of these, and has concluded that the series converges. I agree with his conclusion.

Fresh merely pointed out that the implication is in the wrong order and he is correct. You are also correct that all steps are reversible, but logically the solution contains a flaw and if I would grade it I would subtract a small point.

 

[murshid_islam]

What I had shown was that for a_{n+1} \leq a_n, n has to be greater than \frac{1}{3}. Since n \geq 1, the reverse implication works and therefore a_{n+1} \leq a_n. I can of course write all of that backward so that a_{n+1} \leq a_n follows from n &gt; \frac{1}{3}.
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[fresh_42]

In the OP's work on determining the latter condition, I believe that all the steps are reversible.

Yes, but they are written in the wrong direction. There is a squaring on the way which is not reversible, at least not always. So additional and unmentioned informations come into play. Such steps are notorious flaws in more complex situations. Better to learn it at those easy examples.

 

[Mark44]

but logically the solution contains a flaw

Yep, I see it.

 

[murshid_islam]

What I had shown was that for a_{n+1} \leq a_n, n has to be greater than \frac{1}{3}. Since n \geq 1, the reverse implication works and therefore a_{n+1} \leq a_n. I can of course write all of that backward so that a_{n+1} \leq a_n follows from n &gt; \frac{1}{3}.
.

Or I can assume the opposite, that is, a_{n+1} &gt; a_n, and show a contradiction.
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[murshid_islam]

Hang on, I have checked the function f(x) = \frac{\sqrt x}{2x + 3} and its derivative. There's a maximum at x = 3/2. The function increases up to x = 3/2 and then it decreases.

Does the Alternating Series Test apply here if the sequence \left\{ \frac{\sqrt n}{2n+3} \right\} isn't monotonically decreasing for all n?
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[Mark44]

Hang on, I have checked the function f(x) = \frac{\sqrt x}{2x + 3} and its derivative. There's a maximum at x = 3/2. The function increases up to x = 3/2 and then it decreases.

Does the Alternating Series Test apply here if the sequence \left\{ \frac{\sqrt n}{2n+3} \right\} isn't monotonically decreasing for all n?
.

Yes. The behavior has to be for "large" n. What happens for a few terms for small values of n doesn't matter.

 

[hutchphd]

Fresh merely pointed out that the implication is in the wrong order and he is correct. You are also correct that all steps are reversible, but logically the solution contains a flaw and if I would grade it I would subtract a small point.

So if he used double arrows all would be copacetic?... So if the infererence is IFF always write the double arrow!

 

[fresh_42]

So if he used double arrows all would be copacetic?... So if the infererence is IFF always write the double arrow!

No. As mentioned, squaring isn't reversible.

If you show $A=B \Longrightarrow 1=1$, then you cannot just write it as $A=B \Longleftrightarrow 1=1$. The logic says: True can follow from False, but False cannot follow from True. Hence you have to write $1=1 \Longrightarrow A=B$ if that is what you want to show.

This means in practice, that you scribble $A=B \Longrightarrow 1=1$, hope that it can be reversed, and if so, finally write down $1=1 \Longrightarrow A=B$. It is a matter of hygiene.

 

[hutchphd]

Yes I somehow missed the discussion. But the sum stipulates n is positive...in a formal proof does that not allow the twin arrow at that step? Or do you notate it differently?

 

[fresh_42]

Yes I somehow missed the discussion. But the sum stipulates n is positive...in a formal proof does that not allow the twin arrow at that step? Or do you notate it differently?

In the example we only consider that part of the root function which is above the $x$-axis, so it is equivalent here. This should be mentioned in a proof at this level. Especially in calculus we often have the situation in which things have to be written backwards: For $\varepsilon$ there is an $N$ ... Nobody knows this $N$ from the start, but you have to start with it and end up with $<\varepsilon$. And those deductions are often not reversible!

For squaring we only have $a< b \Longrightarrow a^2<b^2 \Longrightarrow |a|<|b|$. So if $a,b$ aren't numbers but some complicated expressions, then it is not obvious that $a=|a|$ and $b=|b|$. The OP has asked 'Is this ok?' and the answer is 'Almost.'

Maybe I'm a bit spoiled since I saw $A<B \Longrightarrow AC<BC$ in a thesis where the author didn't care about $C>0$, which wasn't obvious at all, since $C$ was an expression which could have been negative without assuming further properties. Another one claimed a group $G\cong SL(2)$ where it actually was $G\cong PSL(2)$. No big deal, because $G$ was a tensor group and of course we have $x \otimes y = \gamma x \otimes (1/\gamma) y$ only up to the center.

Since then I find that it is important to practice such subtleties in time. Not, that I wouldn't make such mistakes myself, too, from time to time.

 

[murshid_islam]

I think it would be better if I consider the function f(x) = \frac{\sqrt x}{2x + 3} and show that f&#039;(x)&lt;0 (and therefore, f(x) is decreasing) on the interval \left( \frac{3}{2}, \infty \right). And so, \left\{\frac{\sqrt n}{2n + 3}\right\} is a decreasing sequence.
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