# Maths Question

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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!

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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#2

(Original post by

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!

**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!

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#3

(Original post by

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!

**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!

Edit: This will work

Last edited by RogerOxon; 1 month ago

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#4

**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!

You should be able to convince yourself that 1/2+1/3+...+1/1000 < (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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(Original post by

First term = 1, next two terms < 1, next four terms ...

**mqb2766**)First term = 1, next two terms < 1, next four terms ...

Sorry if I'm being really silly once again,

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(Original post by

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

**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

Sorry if I'm being really silly once again,

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(Original post by

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 < (because 1/2 < x for x between 1 and 2, 1/3 < x for x between 2 and 3, ...).

Then just evaluate the integral.

**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 < (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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#9

(Original post by

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,

**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,

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.

Last edited by mqb2766; 1 month ago

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#10

**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,

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#11

(Original post by

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,

**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,

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.

Last edited by RogerOxon; 1 month ago

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#12

(Original post by

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.

**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.

[Not saying this is better - but I think it's good to see two slightly different ways of doing this].

Last edited by DFranklin; 1 month ago

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(Original post by

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.

**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.

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(Original post by

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.

**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.

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(Original post by

Basically, you round-up each term to the next power of two:

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.

**RogerOxon**)Basically, you round-up each term to the next power of two:

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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(Original post by

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].

**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].

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(Original post by

Is this A level maths?

**lilbabypenguin**)Is this A level maths?

Sorry if that doesn't help.

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#18

(Original post by

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.

**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.

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#19

(Original post by

It’s ok, I was just thinking you could use sums of geometric series to work this out

**lilbabypenguin**)It’s ok, I was just thinking you could use sums of geometric series to work this out

https://mathworld.wolfram.com/HarmonicSeries.html

Last edited by mqb2766; 1 month ago

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#20

**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.

Last edited by mqb2766; 1 month ago

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