Question

Find An given that the sum of An from 2 to N = 2-1/N

Answer 1

\[\sum_{n=2}^{N} a _{n} = 2-\frac{ 1 }{ N }\]

Answer 2

converges to 2

Answer 3

O.o Are you sure that's all of the question? From first glance, it seems plausible that there could be more than one solution to \[a _{n}\]

Answer 4

That's the whole question

Answer 5

maybe we can find a pattern in the first few terms

Answer 6

we can a2, a3,a4, and then a5
and see if there is some pattern

Answer 7

\[\sum_{n=2}^{2} a_n=2-\frac{1}{2} \\ a_2=2-\frac{1}{2} \\ a_2=\frac{3}{2}\]

Answer 8

\[\sum_{n=2}^{3}a_n=2-\frac{1}{3} \\ a_2+a_3=2-\frac{1}{3}\]
and so on...

Answer 9

\[a_{N}= \sum_{n=2}^{N}a_{n}-\sum_{n=2}^{N-1}a_{n} \]

Answer 10

So, I'm getting 3/2, 5/3, 7/4, 9/5, and so on

Answer 11

So the numerator is increasing by 2 each time, while the denominator is only increasing by 1

Answer 12

maybe I'm doing something wrong but a_3 I got 1/6
I have to resume this later
bbl

Answer 13

ok

Answer 14

@Bobo-i-bo How did you get that?

Answer 15

\[\sum_{n=2}^{N}a_{n}-\sum_{n=2}^{N}a_{n} = (a_2+a_3+...+a_{N-1}+ a_N)-(a_2+a_3+...+a_{N-1})=a_N\]

Answer 16

Oops the limit of the second summand should be N-1 and not N

Answer 17

Is that enough for you to progress? :)

Answer 18

Ok, I see what you did there, but how would that equal to 2-1/N?

Answer 19

It doesn't. But\[\sum_{n=2}^{N}a_n=2-\frac{1}{N}\] and \[\sum_{n=2}^{N-1}a_n=2-\frac{1}{N-1}\] So if you substitute those into the equation I've made...

Answer 20

I'm getting \[\frac{ 1 }{ N(N-1) }\] if I substitute those values into the equation you made

Answer 21

That's what i got too. So \[a_N=\frac{1}{N(N-1)}\] In other words, \[a_n=\frac{1}{n(n-1)}\] which should be the answer. Yet if i try and verify, it doesn't work... so I don't know what's gone wrong '~'

Answer 22

oh, ok

Answer 23

Lol, i'm not trying to be arrogant or anything, but because of what we've done above, I believe that either the question is wrong or else it's a question which is deliberately meant to trick you as there is no solution...

Answer 24

Because I believe my method would constitute a proper proof which lead to a contradiction. (But it may be that i've just gone wrong somewhere in the proof)

Answer 25

Haha, I understand. It's just that I took an Honors Calc 2 class in college and our professor gives us a set of insanely hard problems each week. I was having trouble figuring out how to even start doing this problem.

Answer 26

Sounds fun :P