Question

Formula for this sequence, and explain your answer.
Likewise, does this converge or diverge if we take the sum from 1 to infinity? If it converges, what does it converge to?
{ 1, -1/2, -1/3, 1/4, 1/5, -1/6, -1/7, 1/8, 1/9, -1/10, -1/11, 1/12, 1/13...}

Answer 1

@DollyAcquah

Answer 2

sorry i really this type of math :(

Answer 3

That's ok :)

Answer 4

I know the formula for alternating fractions, but not the one where it alternates every 2

Answer 5

@Hero

Answer 6

I believe that taking the sum from 1 to infinity instead of 0 to infinity might mean the formula is something/n. I know that (-1)^n has to be in there somewhere, too

Answer 7

@ganeshie8

Answer 8

I can't see a way to write an expression for the \(n\)th term without using imaginary numbers. That alternating pattern is tricky ... something like
\[\large a_n=\frac{|i^n|}{n}\]

Answer 9

I thought one might have to use imaginary numbers, since -1^n isn't enough

Answer 10

I suppose there's nothing stopping you from writing \(a_n\) as a piecewise function that considers even and odd \(n\) separately.

Answer 11

this is the answer wolframalpha gave:

a_n = ((1/2+i/2) i^n (-i+(-1)^n))/n

Answer 12

I think the formula I suggested is simpler... No telling what WA did in its calculations.

As for the sum part of the question, you can try comparing to the series \(\displaystyle \sum\dfrac{(-1)^n}{n}\) if it's convergent, or maybe \(\displaystyle \sum\dfrac{1}{n}\) if it diverges.

Answer 13

I eventually had to give up and just have the answer

Answer 14

I tested what WA gave with n=10 -> n = 15, and it holds true. Putting that eqation in to try to look for simpler ones gave me ... i^n * e^(i*pi*n)

Answer 15

WA says it converges to 0, compared to (-1)^n/n, which converges to -log2

Answer 16

You mean the summation, right?

Answer 17

yes

Answer 18

Strange... If you use \(\dfrac{|i^n|}{n}\), the series diverges, but it gives the same sequence.

Answer 19

For the above equation, only positive fractions are outputs...

Answer 20

Oh right, what was I thinking... The absolute value gets rid of the negatives. I meant the real part of \(i^n\).

Answer 21

Wait, that's not right either... We'll need an expression for the sign of \(i^n\).

Answer 22

The one you gave gives:

{ i, -1/2, -i/3, 1/4, i/5, .... }

Answer 23

Well, at any rate you have a closed form for \(a_n\) that does work thanks to WA. How one would get is beyond me :/

Answer 24

thanks for the help and input

Answer 25

funny to see a math genius like you being boggled.

Answer 26

Hardly a genius ;)

Answer 27

by the way, for kicks, I looked at what WA gave for alternating fractions for every 3 terms...and it uses sine, i, and pi :P

Answer 28

*actually, no i. maybe that one would be easier, I don't know. Thanks again.

Answer 29

No problem