Sum (1 Viewer)

[Carrotsticks]

Find the value of the sum:



Assuming the behaviour of the summands is consistent over large n.

 

[viraj30]

is it 1? lol i am in year 11

 

[iSplicer]

Can I use wolfram?

 

[Carrotsticks]

And of course like always, I'm looking for a correct method, not the answer.

 

[bleakarcher]

k/(k+1)!=1/k! - 1/(k+1)!
Now just use a telescoping series to simplify sum and obtain 1 - 1/(n+1)!
The limit of this is 1.

 

[Carrotsticks]

Ding ding ding.

 

[iSplicer]

And of course like always, I'm looking for a correct method, not the answer.

Find the value of the sum:

problem, officer?

 

[lolcakes52]

I had a bit of fun with this using taylor series.



so basically



differentiating both sides we get



let x equal one and bam we get the limit, the advantage of this approach is we could use it to find other sums with powers of x etc.

 

[Carrotsticks]

I had a bit of fun with this using taylor series.



so basically



differentiating both sides we get



let x equal one and bam we get the limit, the advantage of this approach is we could use it to find other sums with powers of x etc.

Just saw this post.

You have to be very careful about differentiating or integrating infinite terms.

Reason is because we can't know for sure (without proof) that the behaviour of the curve (differentiability-wise) is consistent over infinitely many terms.

The power series expansion you had there converges uniformly to e^x, so that gives us free reign to do all sorts of things with it, which happens to include differentiating (also integrating) an infinite number of terms.

 

[abecina]

Just saw this post.

You have to be very careful about differentiating or integrating infinite terms.

Reason is because we can't know for sure (without proof) that the behaviour of the curve (differentiability-wise) is consistent over infinitely many terms.

The power series expansion you had there converges uniformly to e^x, so that gives us free reign to do all sorts of things with it, which happens to include differentiating (also integrating) an infinite number of terms.

Therse are really helpful comments that you add, Carrotsticks. Thanks