comparison test for this series

[tough reef]

still dont know how to do comparison test 💔 idek where to start

[subtle glen]

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[proper solstice]

either set Un=(n/(2n+5)^n then calculate the limit of Un+1/Un if the limit is <1 then the series converges if it is >1 then it diverges

you can also use asomptotic comparaison (n/2n+5)^n=(1/2)^n *(1-5/(2n+5))^n and (1-5/(2n+5))^n ~e^-5 for n approaching infinity so finally Un~(1/2)^n x e^(-5) so via asomptatic comparaison to the geometric sequence ( the series being convergent here) the series of Un converges as well

[tough reef]

sorry i don't understand this part: (n/2n+5)^n=(1/2)^n *(1-5/(2n+5))^n and (1-5/(2n+5))^n ~e^-5

[proper solstice]

my bad I may have been a bit hasty while typing basically the limit of (1-5/(2n+5))^n when n approaches infinity is e^-5 this stems from the fact that nln(1-5/(2n+5)) converges to -5 (limit of rate of change or taylor expansion can prove it if you what those are) and so applying the exponatial via continuity the timit of (1-5/(2n+5))^n is e^-5

[tough reef]

so what about the geometric sequence part? and (1/2)^n ?

[proper solstice]

Ok basically the idea here is to use this :
If (Un) and (Vn) are two positive sequences such that Un~Vn for n approaching infinity then if the series of Vn converges so does that of Un

Here the geometric series converges

The series of (1/2)^n * e^(-5) converges because 1/2<1

So your sequence (Un) being equivalent to (1/2)^n *e^-5 as n approaches infinity means that the series of Un converges

I don’t think it tends to exp something I find it indeed converges

[mighty junco]

No the general term tends to 0

You can see that its log is equivalent to n log(1/2)

[mighty junco]

This is in fact a pretty good guess imo

You can upper bound the n/(2n+5) by n / (2n)

Since you just remove the 5 in the denominator

Now since this is nonnegative, exponentiation to the power n conserve orders

Hence (n / (2n+5))^n <= (1/2)^n

And this is a fairly simple guess that leverages your intuition about the geometric series

[drowsy steppe]

oh wait you're right

mornings

[mighty junco]

Happens man, no worries

[tough reef]

i'm very lost 💔

[mighty junco]

What's got you lost?

[tough reef]

all of it... i know for the comparison test you have to find a simpler version of the series and see if that converges or diverges, and that should tell you if the original series converges or diverges, but i don't see where you did that in your instructions

[mighty junco]

well

suppose a_n and b_n are all positive

if b_n is bigger and the series of b_n converges

then you have a conclusion about the series of a_n right?

[tough reef]

right

[mighty junco]

so all I did was find a general term that is slightly bigger

by taking a smaller denominator

n / (2n + 5) <= n / (2n) = 1/2

so (n / (2n+5))^n <= (1/2)^n

but you know well that the series of (1/2)^n converges

[tough reef]

ohhhhh

so (n / (2n+5))^n converges

[mighty junco]

the series of (n / (2n+5))^n converges yes

[tough reef]

awesome i understand now

thank u so much

[mighty junco]

np, glad it was helpful

[tough reef]

+close