20 points! no idea what to

find \lim_ {n \to \infty} a_n[1+ (-1)^n/n]

!

favorite 2 limn→∞∫∞0sin(x/n)(1+x/n)ndx

I've been able to show that the integral is bounded above by 1 (several ways). One of the simplest is just letting u=x/n

and doing some bounding above. It must converge to something (if it is monotone increasing.. don't think so), but all my efforts to use Lebesgue Dominated Convergence have failed. I can't seem to dominate this function...

However, if I could dominate it the integrand converges pointwise to 0 and so then the integral would be 0, which I think is incorrect.

I have found it sufficient to deal with

limn→∞∫n0sin(x/n)(1+x/n)ndx

I tried to turn it into a series and use the converse of the integral test, but that hasn't went well.

I've used Egoroff's theorem to try to use uniform convergence but that failed.

Some of the inequalities and ideas I've tried to use in a solution include:

sin(x/n)≤x/n

after a u

sub I've tried to turn it into a riemann sum... failed

(1+x/n)n≥1+x

and the binomial theorem

(1+x/n)n

converges to ex and it is increasing in n so e−x≤1(1+x/n)n

I've tried to use that the integrand is measurable and subtract out different measurable sets from the integral to make it more convenient..

I've been playing with this on and off for a month or so and no luck. I appreciate any help here.

Update: From mixedmath's suggestion

splitting it into [0,10]

and [10,∞] it is clear that the integrand is dominated on [0,10]

(say by 1) and so lebesgue sends that portion to 0.

On the larger interval I am still running into a problem making it bounded nicely. If I try a u

sub the bounds get messy. I would like to choose an n

so that

2e−x≤(1+x/n)n

for all x∈[10,∞]

.. maybe:

there exists an N

such that for all n≥N

e10/2<(1+10/n)n

but for x>10

..

e10/2<(1+10/n)n<(1+x/n)n

but that's not quite right..

Aye bruh i don’t know this but someone else who smarter than us will answer this

The answer is d 18ft