# Integrals

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Student 999

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#1

Stuck on the deduce part of showing the inequality contains the sum of 1/n from 2 to N

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tonyiptony

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(Original post by

Stuck on the deduce part of showing the inequality contains the sum of 1/n from 2 to N

**Student 999**)Stuck on the deduce part of showing the inequality contains the sum of 1/n from 2 to N

(2) You can add inequalities. i.e. if a>b and c>d, then a+c>b+d.

I'm assuming you're stuck at the start. If you know these facts, the deduction should be fairly straightforward.

In fact, it's kind of a common "trick" involving sums and inequalities.

EDIT: Here's what you can do:

Spoiler:

Show

I guess I just said it in a very convoluted way that you can slap the summation sign on all parts of the inequality.

Last edited by tonyiptony; 1 month ago

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Student 999

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#3

I'm struggling to see where I've gone wrong

(Original post by

(1) You can plug any number in n, i.e. n=2,3,4... etc., and the inequality is still true.

(2) You can add inequalities. i.e. if a>b and c>d, then a+c>b+d.

I'm assuming you're stuck at the start. If you know these facts, the deduction should be fairly straightforward.

In fact, it's kind of a common "trick" involving sums and inequalities.

EDIT: Here's what you can do:

**tonyiptony**)(1) You can plug any number in n, i.e. n=2,3,4... etc., and the inequality is still true.

(2) You can add inequalities. i.e. if a>b and c>d, then a+c>b+d.

I'm assuming you're stuck at the start. If you know these facts, the deduction should be fairly straightforward.

In fact, it's kind of a common "trick" involving sums and inequalities.

EDIT: Here's what you can do:

Spoiler:

Show

I guess I just said it in a very convoluted way that you can slap the summation sign on all parts of the inequality.

Last edited by Student 999; 1 month ago

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mqb2766

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#4

What do you think your problem is? Is it the 1-1/N on the right rather than 1?

Last edited by mqb2766; 1 month ago

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Student 999

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#5

(Original post by

What do you think your problem is? Is it the 1-1/N on the right rather than 1?

**mqb2766**)What do you think your problem is? Is it the 1-1/N on the right rather than 1?

But it isn't stated that N is a big number though?

The only other alternative that I can think of is to use 1-1/N< 1 but then the inequality signs will be different if you sub it in

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mqb2766

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(Original post by

Yes, are you going to say that N tends to infinity 1-1/N tends to 1 to get that form.

But it isn't stated that N is a big number though?

The only other alternative that I can think of is to use 1-1/N< 1 but then the inequality signs will be different if you sub it in

**Student 999**)Yes, are you going to say that N tends to infinity 1-1/N tends to 1 to get that form.

But it isn't stated that N is a big number though?

The only other alternative that I can think of is to use 1-1/N< 1 but then the inequality signs will be different if you sub it in

log()-1 < sum()

so you can conclude that the sum -> inf as log ->inf.

The values of 0 and 1 in the inequality are largely irrelevant as all you want to say is

log() ~< sum()

and as long as the left increasses to infinity, any small corrections to the sum get lost in the noise and we make the same conclusion that the sum() increases to infinitiy.

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tonyiptony

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**Student 999**)

Yes, are you going to say that N tends to infinity 1-1/N tends to 1 to get that form.

But it isn't stated that N is a big number though?

The only other alternative that I can think of is to use 1-1/N< 1 but then the inequality signs will be different if you sub it in

At this point of the question, we didn't mention n being "large enough". It's just that n>1 (presumably a natural number as well, that's kind of the unspoken rule of math). So a simple will do.

Unless you've jumped ahead and I'm not aware.

Last edited by tonyiptony; 1 month ago

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#8

I'm now confused again haha, from my above post I have ln(N) - sum <=1-1/N which isn't what they asked me to deduce. Hence I said that 1-1/N < 1 for all natural numbers of N however that'll result to my inequality to be ln(N) -sum < 1 which is what I'm confused about since I have the inequality sign <1 not <=1 which is what they want me to show.

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(Original post by

I'm now confused again haha, from my above post I have ln(N) - sum <=1-1/N which isn't what they asked me to deduce. Hence I said that 1-1/N < 1 for all natural numbers of N however that'll result to my inequality to be ln(N) -sum < 1 which is what I'm confused about since I have the inequality sign <1 not <=1 which is what they want me to show.

**Student 999**)I'm now confused again haha, from my above post I have ln(N) - sum <=1-1/N which isn't what they asked me to deduce. Hence I said that 1-1/N < 1 for all natural numbers of N however that'll result to my inequality to be ln(N) -sum < 1 which is what I'm confused about since I have the inequality sign <1 not <=1 which is what they want me to show.

Last edited by mqb2766; 1 month ago

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