The Student Room Group
Reply 1
What's the series?
Reply 2
It's a longer series. The exact series doesn't really matter.

How about I just write:

Sigma[n]En = lim[T-->∞]Sigma[n]En

And then solve for the limit?
Reply 3
Surely arithmetic series diverge?
Siddhartha
Ok, I'm just slightly confused.
I have to find the sum to infinity of a series, which turns out to be arithmetic in its nature. Is it correct to find the sum from n=1 to n=n, because you clearly cannot define n=infinity.


You cant define n=infinity, but you can consider the limit as n tends to infinity.

Also, as aleady said, an arithmetic progression diverges since its comparable to the sum of n, which is divergent. The "sum to infinity" is only really heard of in geometric series' in my experience.
Reply 5
mr_jr
`in my experience.
That's probably because you don't have much experience of convergent series other than geometric ones.

But you can use "sum to infinity" for any series that converges; for example 11n(n+1)=1\sum_1^\infty \frac{1}{n(n+1)} = 1
Reply 6
Original post by DFranklin
That's probably because you don't have much experience of convergent series other than geometric ones.

But you can use "sum to infinity" for any series that converges; for example 11n(n+1)=1\sum_1^\infty \frac{1}{n(n+1)} = 1


Sorry can someone explain this further
The sum of all the positive consecutive integers to infinity is equal to -1/12.

Just this this was interesting to share :smile:
Original post by Y11_Maths
The sum of all the positive consecutive integers to infinity is equal to -1/12.

Just this this was interesting to share :smile:


Hope you're joking.
Reply 9
Original post by Y11_Maths
The sum of all the positive consecutive integers to infinity is equal to -1/12.

Just this this was interesting to share :smile:

It's not really. The Numberphile video isn't one of their best - this explains things much better, although it's quite long!

Original post by Baeha
Sorry can someone explain this further

Please start a new thread. This one is ancient.