maya_jai_singh
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Does anyone have any idea how to prove 1 + 1/2 + 1/3 + ... + 1/1000 < 10?

I have no idea where to start with this, so even just a hint or a first step would be greatly appreciated.

Thank you!
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mqb2766
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(Original post by maya_jai_singh)
Does anyone have any idea how to prove 1 + 1/2 + 1/3 + ... + 1/1000 < 10?

I have no idea where to start with this, so even just a hint or a first step would be greatly appreciated.

Thank you!
First term = 1, next two terms < 1, next four terms ...
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RogerOxon
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(Original post by maya_jai_singh)
Does anyone have any idea how to prove 1 1/2 1/3 ... 1/1000 < 10?

I have no idea where to start with this, so even just a hint or a first step would be greatly appreciated.

Thank you!
I've not really thought about this, but I'd try increasing some of the terms to make the sum simpler. I'd start with powers of 2, so 1 1/2 1/2 1/4 1/4 1/4 1/4 1/8 ..

Edit: This will work
Last edited by RogerOxon; 1 month ago
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DFranklin
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(Original post by maya_jai_singh)
Does anyone have any idea how to prove 1 + 1/2 + 1/3 + ... + 1/1000 < 10?

I have no idea where to start with this, so even just a hint or a first step would be greatly appreciated.

Thank you!
The method already suggested is fine and the one I would choose, but if you want more "strings to your bow":

You should be able to convince yourself that 1/2+1/3+...+1/1000 < \int_1^{1000} \frac{1}{x}\,dx (because 1/2 < x for x between 1 and 2, 1/3 < x for x between 2 and 3, ...).

Then just evaluate the integral.
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maya_jai_singh
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(Original post by mqb2766)
First term = 1, next two terms < 1, next four terms ...
Apologies if I'm being slightly thick (my maths isn't that great - really sorry) but I'm still slightly stuck. Pls do you mind going one step further, perhaps proving it if you have time just so I could use a reference - no worries if not!

Sorry if I'm being really silly once again,
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maya_jai_singh
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(Original post by RogerOxon)
I've not really thought about this, but I'd try increasing some of the terms to make the sum simpler. I'd start with powers of 2, so 1 1/2 1/2 1/4 1/4 1/4 1/4 1/8 ..

Edit: This will work
Apologies if I'm being slightly thick (my maths isn't that great - really sorry) but I'm still slightly stuck. Pls do you mind going one step further, perhaps proving it using this method if you have time just so I could use a reference - no worries if not!

Sorry if I'm being really silly once again,
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maya_jai_singh
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(Original post by DFranklin)
The method already suggested is fine and the one I would choose, but if you want more "strings to your bow":

You should be able to convince yourself that 1/2+1/3+...+1/1000 < \int_1^{1000} \frac{1}{x}\,dx (because 1/2 < x for x between 1 and 2, 1/3 < x for x between 2 and 3, ...).

Then just evaluate the integral.
This makes sense. I get why the inequality works now however never in a million years would I have spotted this myself! Thanks so much for your time,
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username5490332
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Is this A level maths?
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mqb2766
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(Original post by maya_jai_singh)
Apologies if I'm being slightly thick (my maths isn't that great - really sorry) but I'm still slightly stuck. Pls do you mind going one step further, perhaps proving it if you have time just so I could use a reference - no worries if not!

Sorry if I'm being really silly once again,
Just write the first 16 terms (say) out and group them so that the sum of each group is <= 1.
What do you notice about
* The first element in each group
* The number of elements in each group.
generalise the pattern to 1000 terms.

upload what you've tried.
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DFranklin
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(Original post by maya_jai_singh)
Apologies if I'm being slightly thick (my maths isn't that great - really sorry) but I'm still slightly stuck. Pls do you mind going one step further, perhaps proving it if you have time just so I could use a reference - no worries if not!

Sorry if I'm being really silly once again,
I don't want to tread on mqb's toes answering this, but just to say: this is actually quite a tough thing to spot (but easy once you've seen it), so don't be too hard on yourself if you're finding it tricky.
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RogerOxon
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(Original post by maya_jai_singh)
Apologies if I'm being slightly thick (my maths isn't that great - really sorry) but I'm still slightly stuck. Pls do you mind going one step further, perhaps proving it using this method if you have time just so I could use a reference - no worries if not!

Sorry if I'm being really silly once again,
Basically, you round-up each term to the next power of two:
1+ \frac{1}{2}+ \frac{1}{3}+ \frac{1}{4}+ \frac{1}{5}+ \frac{1}{6}+ \frac{1}{7}+ \frac{1}{8}+ \frac{1}{9}+.. &lt; 1+ \frac{1}{2}+  \frac{1}{2}+ \frac{1}{4}+ \frac{1}{4}+ \frac{1}{4}+ \frac{1}{4}+ \frac{1}{8}+ \frac{1}{8}+..

The first term is 1
The next two terms sum to less than 1 (1 in the rounded expression)
The next four terms sum to less than 1 (1 in the rounded expression)
The next eight terms sum to less than 1 (1 in the rounded expression)
So, we have 1+2+4+8+16+32+64+128+256+489 (the last is 1000 - 511 terms, the sum is 1 for the last 512 rounded-up terms, but that's too many). The first group sums to 1. All the others sum to less than 1. As we have 10 groups, we have shown that the sum is less than 10.
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DFranklin
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(Original post by RogerOxon)
So, we have 1+2+4+8+16+32+64+128+256+489 (the last is 1000 - 511 terms, the sum is 1 for the last 512 rounded-up terms, but that's too many). The first group sums to 1. All the others sum to less than 1. As we have 10 groups, we have shown that the sum is less than 10.
Alternatively, you could prove S = 1+1/2+1/3+...1/1023 <10 (thereby avoiding having to truncate the last group) and then obviously 1+1/2+...+1/1000 < S < 10.

[Not saying this is better - but I think it's good to see two slightly different ways of doing this].
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maya_jai_singh
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(Original post by mqb2766)
Just write the first 16 terms (say) out and group them so that the sum of each group is <= 1.
What do you notice about
* The first element in each group
* The number of elements in each group.
generalise the pattern to 1000 terms.

upload what you've tried.
Ahh! I see it now, that's wonderful. Thank you so much for your time and efforts, greatly appreciated.
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maya_jai_singh
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(Original post by DFranklin)
I don't want to tread on mqb's toes answering this, but just to say: this is actually quite a tough thing to spot (but easy once you've seen it), so don't be too hard on yourself if you're finding it tricky.
Thank you so much for this. It's rather reassuring, thanks for your time and efforts too it does mean a lot!
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maya_jai_singh
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(Original post by RogerOxon)
Basically, you round-up each term to the next power of two:
1+ \frac{1}{2}+ \frac{1}{3}+ \frac{1}{4}+ \frac{1}{5}+ \frac{1}{6}+ \frac{1}{7}+ \frac{1}{8}+ \frac{1}{9}+.. &lt; 1+ \frac{1}{2}+  \frac{1}{2}+ \frac{1}{4}+ \frac{1}{4}+ \frac{1}{4}+ \frac{1}{4}+ \frac{1}{8}+ \frac{1}{8}+..

The first term is 1
The next two terms sum to less than 1 (1 in the rounded expression)
The next four terms sum to less than 1 (1 in the rounded expression)
The next eight terms sum to less than 1 (1 in the rounded expression)
So, we have 1+2+4+8+16+32+64+128+256+489 (the last is 1000 - 511 terms, the sum is 1 for the last 512 rounded-up terms, but that's too many). The first group sums to 1. All the others sum to less than 1. As we have 10 groups, we have shown that the sum is less than 10.
This makes so much sense! Thanks so much for your time and efforts, greatly appreciated!
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maya_jai_singh
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(Original post by DFranklin)
Alternatively, you could prove S = 1+1/2+1/3+...1/1023 <10 (thereby avoiding having to truncate the last group) and then obviously 1+1/2+...+1/1000 < S < 10.

[Not saying this is better - but I think it's good to see two slightly different ways of doing this].
Yeah, this also makes sense. Thanks!
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maya_jai_singh
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(Original post by lilbabypenguin)
Is this A level maths?
I haven't really gotten this from an A level maths based website, it's designed more towards trickier questions one may find in an admission test or get asked at an interview.

Sorry if that doesn't help.
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username5490332
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(Original post by maya_jai_singh)
I haven't really gotten this from an A level maths based website, it's designed more towards trickier questions one may find in an admission test or get asked at an interview.

Sorry if that doesn't help.
It’s ok, I was just thinking you could use sums of geometric series to work this out
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mqb2766
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(Original post by lilbabypenguin)
It’s ok, I was just thinking you could use sums of geometric series to work this out
It's a harmonic series, rather than geometric.
https://mathworld.wolfram.com/HarmonicSeries.html
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mqb2766
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(Original post by maya_jai_singh)
I haven't really gotten this from an A level maths based website, it's designed more towards trickier questions one may find in an admission test or get asked at an interview.

Sorry if that doesn't help.
If you're practicing for an interview, I'd concentrate more on discussing what your existing ideas are, rather than partially learning new techniques from question banks, which your interviewer is likely to be aware of. Even if it's wrong, it's important to meet people halfway and describe what you think, rather than just saying "I don't understand".
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