infinite series problem .:Resolved:.

[voidflux]

Hello everyone, I posted this question before, but i said i resolved it becuase I figured it out. But I discovered i'm not allowed to use the integral test to prove

Infinity
E         1/n
n = 1

diverges.

Note: E stands for sumnation
Anyone know how ur suppose to prove this diverges without the integral test?
Thanks.

[voidflux]

that does make sense! thanks alot!

[TheManWhoCan]

I might be able to help you with the expressing that as a proof thing:

Consider S = 1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + ...

Now 1/2 + 1/3 < 1, so

S > 1 + 1 + 1/4 + 1/5 + 1/6 + 1/7 + ...

Also, 1/4 + 1/5 + 1/6 + 1/7 < 1:

S > 1 + 1 + 1 + ...

You can keep doing that forever and a day, i.e. the sum of the next 8 terms will be less than 1, then the sum of the 16 terms after that will be less than 1. That can be proved quite easily (so I can't be arsed to reproduce it). But the end result is you're gonna be adding an infinite amount of 1's together and still not get S, hence the series diverges and S is undefined.

[Guv]

TheManWhoCan: You remembered the proof incorrectly. Your proof only proves that the sum is less than an infinite sum of ones. None of your groups is greater than one

The grouping you need is the following.

• 1
• 1/2
• 1/3 + 1/4, which is > 2/4
• 1/5 + 1/6 + 1/7 + 1/8, which is > 4/8
• 1/9 + 1/10 . . . + 1/16, which is > 8/16

Each group ends with 1/2ⁿ. Each group is greater than 1/2 (except the second which is = 1/2)

Hence the sum is greater than 1/2 + 1/2 + 1/2 . . . + 1/2 . . .

[TheManWhoCan]

Yeh well it's a while since I've seen it