simple series question Watch

thorn0123
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  \displaystyle \sum_{n=1}^\infty (-1)^{n-1}a_n where  (a_n) is a monotone decreasing sequence of nonnegative numbers with  a_n \rightarrow 0 by the alternating series test, series of this form always converge.

Show that  0 \leq \displaystyle\sum_{n=1}^\infty (-1)^{n-1}a_n \leq a_1

There is a hint in the question:

if  (S_N) denotes the sequence of partial sums, consider the subsequences  (S_{2N}) and  (S_{2N-1}) can you show that one is decreasing while the other is increasing?

Well, firstly, I don't understand what this hint is getting at, if I could show that one is increasing while the other is decreasing then I wouldn't know how to proceed from there. Despite that, I cannot even show one is decreasing while the other is increasing, here is my attempt:

 S_N = \displaystyle \sum_{n=1}^N (-1)^{n-1} a_n = \sum_{n=1}^N(-1)^{2n-1}a_{2n} + \sum_{n=1}^N(-1)^{2n} a_{2n+1} = \sum_{n=1}^N (a_{2n+1} - a_{2n}) since  a_{2n} \geq a_{2n+1}  S_N \leq 0 (I find this rather strange so I'm assuming ive gone wrong already)

now,  S_{2N} = \displaystyle \sum_{n=1}^{2N} (a_{2n+1} - a_{2n}) \geq S_{2(N+1)} so  S_{2N} is decreasing

also,  S_{2N-1} = \displaystyle \sum_{n=1}^{2N-1} (a_{2n+1} - a_{2n}) \geq S_{2(N+1)-1} so  S_{2N-1} is also decreasing

now I didn't even manage to solve the hint x_x,

any help is appreciated - thanks.
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Shillington
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(Original post by thorn0123)
Let  S = \displaystyle \sum_{n=1}^\infty (-1)^{n-1}a_n where  (a_n) is a monotone decreasing sequence of nonnegative numbers with  a_n \rightarrow 0 by the alternating series test, series of this form always converge.

Show that  0 \leq S \leq a_1

There is a hint in the question:

if  (S_N) denotes the sequence of partial sums, consider the subsequences  (S_{2N}) and  (S_{2N-1}) can you show that one is decreasing while the other is increasing?
Showing the hint is true:

First of all, don't get sloppy with your notation. You've got Ns that should be infinities and an n=1 that should be an n=0.

Now. Don't bother writing out any sigmas, just follow the logic:

a_n is a monotone decreasing sequence of non negative numbers with a_n \to 0. An example sequence of this is a_n=\frac{1}{n}. Now, the summation is a summation of the following sequence:

(-1)^{n-1}a_n

Now, try out what's happening with S_2N and S_2N+1 with this sequence to try and get a feel for how to show the hint is true for the arbitrary sequence.


Spoiler:
Show
The trick to showing that S_2N is increasing is to show that S_{2N}=S_{2N-2}+(a_{2N-1}-a_{2N})\geq S_{2N-2}. So if you can show that a_{2N-1}-a_{2N}\geq 0 then you're done!

Using the hint:

We know that the  lim S_N =  lim S_{2N} = lim S_{2N+1} (by easy results). Either we will always have S_2N>S_2N+1 or the other way around. As one of these sequences is decreasing and the other is increasing, we know that S_N will be sandwiched between the two (for some appropriate choice of N_0 lets say).

Can you fill in the details?
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thorn0123
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Hi, first of all thank for you for your reply.

(Original post by Shillington)
Showing the hint is true:

First of all, don't get sloppy with your notation. You've got Ns that should be infinities and an n=1 that should be an n=0.
Could you please specify where? I looked over it again and I don't see where there should be a n = 0 and an infinity sign instead of N. S_N denotes the partial sums. edit: edited my first post to make it more clear

(Original post by Shillington)
Showing the hint is true:
Now. Don't bother writing out any sigmas, just follow the logic:

a_n is a monotone decreasing sequence of non negative numbers with a_n \to 0. An example sequence of this is a_n=\frac{1}{n}. Now, the summation is a summation of the following sequence:

(-1)^{n-1}a_n

Now, try out what's happening with S_2N and S_2N+1 with this sequence to try and get a feel for how to show the hint is true for the arbitrary sequence.
I tried that sequence  a_n = 1/n to start with.  S_{2N} = \displaystyle \sum_{n=1}^{2N}(-1)^{n-1}\dfrac{1}{n} right (for a_n = 1/n)? I don't see how this could be increase, while S_{2N-1} could possible be decreasing. To me, S_{2N} is summing to 2N while S_{2N-1} is summing to one term less - I could see it if we replace "n" by "2n" but that's not what the question says.
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thorn0123
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(Original post by ztibor)
.
(Original post by ghostwalker)
.
Anyone ?
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Shillington
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(Original post by thorn0123)
Hi, first of all thank for you for your reply.


Could you please specify where? I looked over it again and I don't see where there should be a n = 0 and an infinity sign instead of N. S_N denotes the partial sums. edit: edited my first post to make it more clear
You're right here, there shouldn't be infiinites, but the Ns are all over the place. On the RHS you've got a_2N terms and on the left you only go up to a_N. On the LHS you've got an a_1 term, on the right you have no term. The long LaTeX line is clearly incorrect.


I tried that sequence  a_n = 1/n to start with.  S_{2N} = \displaystyle \sum_{n=1}^N (-1)^{n-1}\dfrac{1}{n} right (for a_n = 1/n)? I don't see how this could be increase, while S_{2N-1} could possible be decreasing. To me, S_{2N} is summing to 2N while S_{2N-1} is summing to one term less - I could see it if we replace "n" by "2n" but that's not what the question says.
What are the first few terms?

S_1=1
S_3=5/6 (=0.8333333...)
S_5=0.7833333..

S_2=0.5
S_4=0.5833333
S_6=0.616666...

When you say you actually tried this first, what did you mean?

Finally, try and draw a picture of what's happening. I think this is the most intuitive way of seeing what's going on.
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thorn0123
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(Original post by Shillington)
You're right here, there shouldn't be infiinites, but the Ns are all over the place. On the RHS you've got a_2N terms and on the left you only go up to a_N. On the LHS you've got an a_1 term, on the right you have no term. The long LaTeX line is clearly incorrect.



What are the first few terms?

S_1=1
S_3=5/6 (=0.8333333...)
S_5=0.7833333..

S_2=0.5
S_4=0.5833333
S_6=0.616666...

When you say you actually tried this first, what did you mean?

Finally, try and draw a picture of what's happening. I think this is the most intuitive way of seeing what's going on.
Sorry, I just tried to think about it really without writing down anything - your numerical example helped.

One more query to clarify:

(Original post by Shillington)
Spoiler:
Show
The trick to showing that S_2N is increasing is to show that S_{2N}=S_{2N-2}+(a_{2N-1}-a_{2N})\geq S_{2N-2}. So if you can show that a_{2N-1}-a_{2N}\geq 0 then you're done!
How did you get S_{2N}=S_{2N-2}+(a_{2N-1}-a_{2N})\geq S_{2N-2}? Is it because when n is even (e.g. 2N) we get a negative term so  -a_{2N} and when n is odd we get a positive term e..g  a_{2N-1} ?
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Shillington
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(Original post by thorn0123)
How did you get S_{2N}=S_{2N-2}+(a_{2N-1}-a_{2N})\geq S_{2N-2}? Is it because when n is even (e.g. 2N) we get a negative term so  -a_{2N} and when n is odd we get a positive term e..g  a_{2N-1} ?
Yes.
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thorn0123
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(Original post by Shillington)
Yes.
Right,

so I have managed to prove that  S_{2N} is increasing and  S_{2N-1} is decreasing, now I'm trying to put it all together:

heres what I'm thinking so far:

 (S_{2N}) = S_2, S_4, ... S_N, ... , S_{2N}
 (S_{2N-1}) = S_1, S_3, ... S_N, ..., S_{2N-1}

so  S_{2N-1} < S_{N} < S_{2N} but not sure where to really go from here

I have read your first post again and have some questions on it:

(Original post by Shillington)
Either we will always have S_2N>S_2N+1 or the other way around.
I don't understand why you're pointing this out - and is there no way to determine "which way round" it is?

anyhelp on where to go from here,

thanks
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Shillington
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I have read your first post again and have some questions on it:

I don't understand why you're pointing this out - and is there no way to determine "which way round" it is?

anyhelp on where to go from here,

thanks
Covering this first:

The set up we have here is that we have a convergent sequence a_n which has two subsequences a_{2n} and a_{2n+1}. Because a_n converges, both a_2n and a_2n+1 also converge (all to the same number). We know that a_2n is increasing and that a_2n+1 is decreasing (the reason that I said "or the other way around" is that I hadn't bothered to work out which way around it was at the time of posting).

If you had to draw a picture of these three sequences, you'd have a sequence of dots converging to a point. Half of the dots would be converging from below, the other half would be converging from above. So can you see that we will always have that S_2M<S_2N+1 (where I mean to have M and N, as all that matters is the parity).

You can prove this as follows:

Suppose that it doesn't hold for some M and N. Then for all subsequent terms, the odd S's will be strictly below the even S's, hence the two subsequences can't converge to the same number, contradiction. (Obviously I leave you to fill in a few more words around this.)

(Original post by thorn0123)
Right,

so I have managed to prove that  S_{2N} is increasing and  S_{2N-1} is decreasing, now I'm trying to put it all together:

heres what I'm thinking so far:

 (S_{2N}) = S_2, S_4, ... S_N, ... , S_{2N}
 (S_{2N-1}) = S_1, S_3, ... S_N, ..., S_{2N-1}

so but not sure where to really go from here
The line  S_{2N-1} &lt; S_{N} &lt; S_{2N} is not true. If N is odd then the S_N term will be greater than the S_2N+1 term. Have a go at some numerical examples to see that this is true.

Here is a summary of what you know that should let you finish off the question:

S_2N+1>=S_2M for any M and N.

S_1>=S_2N+1 for all N.

S_2<=S_N for all N.

Now you need to show that S_2>=0 and you're done.

Note that you've made this whole question into a bit of a mountain, drawing a decent picture of the sequence (-1)^n+1*a_n would have probably given you more intuition than pushing around notation.
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thorn0123
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(Original post by Shillington)
Covering this first:

The set up we have here is that we have a convergent sequence a_n which has two subsequences a_{2n} and a_{2n+1}. Because a_n converges, both a_2n and a_2n+1 also converge (all to the same number). We know that a_2n is increasing and that a_2n+1 is decreasing (the reason that I said "or the other way around" is that I hadn't bothered to work out which way around it was at the time of posting).

If you had to draw a picture of these three sequences, you'd have a sequence of dots converging to a point. Half of the dots would be converging from below, the other half would be converging from above. So can you see that we will always have that S_2M<S_2N+1 (where I mean to have M and N, as all that matters is the parity).

You can prove this as follows:

Suppose that it doesn't hold for some M and N. Then for all subsequent terms, the odd S's will be strictly below the even S's, hence the two subsequences can't converge to the same number, contradiction. (Obviously I leave you to fill in a few more words around this.)



The line  S_{2N-1} &lt; S_{N} &lt; S_{2N} is not true. If N is odd then the S_N term will be greater than the S_2N+1 term. Have a go at some numerical examples to see that this is true.

Here is a summary of what you know that should let you finish off the question:

S_2N+1>=S_2M for any M and N.

S_1>=S_2N+1 for all N.

S_2<=S_N for all N.

Now you need to show that S_2>=0 and you're done.

Note that you've made this whole question into a bit of a mountain, drawing a decent picture of the sequence (-1)^n+1*a_n would have probably given you more intuition than pushing around notation.
yeah you're right - just drew a diagram and it all become pretty clear.

Appreciate all your help, thank you.
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