Maths Question

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!

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!

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 ...

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

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.

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,

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,

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,

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.

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.

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.

Basically, you round-up each term to the next power of two:
$1+ \frac{1}{2}+ \frac{1}{3}+ \frac{1}{4}+ \frac{1}{5}+ \frac{1}{6}+ \frac{1}{7}+ \frac{1}{8}+ \frac{1}{9}+.. < 1+ \frac{1}{2}+ \frac{1}{2}+ \frac{1}{4}+ \frac{1}{4}+ \frac{1}{4}+ \frac{1}{4}+ \frac{1}{8}+ \frac{1}{8}+..$

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.

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].

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.

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

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.