Why?

[dense beacon]

Why can I use the sum 1/(1+k)^2=sum((n+1)(-1)^n*k^n) for |k|<1? The integral is from 0 to 1 but the domain is less than 1

[sinful temple]

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[barren drum]

"but the domain is less than 1" what does this mean?

@dense beacon

[dense beacon]

I meant that the sum only konverges if |k|<1 but the boundaries of the integral are 0 and 1

So why is it a valid move?

[barren drum]

which sum?

what is the task? to calculate I_1?

[dense beacon]

1/(1+k)^2=sum from n=0 to n=inf of (n+1)(-1)^nk^n for |k|<1

[dense beacon]

Ye but I only wanted to know why I can use the sum even tho it diverges for k=1

[barren drum]

$$\frac{1}{(1+k)^2} = \sum _{n=0}^\infty (-1)^n(n+1)k^n $$

[atomic belfry]

aL

[barren drum]

where is this sum in the picture

here?

[dense beacon]

It's already simplified

[barren drum]

if the integral runs from 0 to 1, then k in (0,1)

confusing notation tbh

[dense beacon]

Sry but I still don't get it

[barren drum]

you said convergence holds for |k|<1

if k in (0,1), then..?

[dense beacon]

Uh

[barren drum]

do you wonder about the equality

[dense beacon]

I wonder why it's a valid move

[barren drum]

$$ \int _0^1 \frac{x^{-1/2}\ln x}{(1+x)^2}dx = \sum _{k=0}^\infty (-1)^k (1+k) \int _0^{1} x^{k-1/2}\ln x dx $$

[atomic belfry]

aL

[barren drum]

is your question why this equality holds?

[dense beacon]

No

If the sum diverges at k=1 why can we use it in the integral?

The integral boundaries are k=0 and k=1

[barren drum]

x in (0,1)

[dense beacon]

What u mean?

[barren drum]

this means

the variable is in (0,1)

it doesn't matter that you get divergence at end point

if it is undefined for any finite number of points, it makes no difference

[orchid girder]

I think the question kinda makes sense given that the argument "x in (0, 1)" isn't all that convincing to show that the equality holds... For instance, when considering the geometric series, it would hold a pathological identity, just under the pretense that $x \in (0, 1)$:
$$\sum_{k=0}^{\infty} (-1)^k \int_{0}^{1} x^k dx = \int_{0}^{1} \frac{dx}{1+x}$$

[atomic belfry]

Rion

[barren drum]

well, alleluja for you for understanding what the question even is

[orchid girder]

That being said I have no clue what can answer this question

[barren drum]

essentially why can we exchange limit and integration?

[orchid girder]

Why we can do it here

[barren drum]

something something uniform convergence

[orchid girder]

and not in the Grandi series

[barren drum]

lemme cook..

[orchid girder]

Also I did a mistake here, this thing indeed converges

[dense beacon]

I think I answered the question myself 😄

Appreciate your help tho

[orchid girder]

I totally forgot about dividing by 1/k when I integrate

[dense beacon]

+close