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Direct comparison test for series convergence

Determine the convergence/divergence of the series from n=2 to infinity
of :

(ln⁡n)e−n (\ln n)e^{-\sqrt n}

i know i can use the direct comparison test, ive done so by letting a function bn=n.e−n b_n = n.e^{-\sqrt n} , my questions are:

1) does it matter what function you use for b(n), as long as they fit the criteria that 0≤an≤bn 0 \leq a_n \leq b_n , and if so is it alright that i've used bn=n.e−n b_n = n.e^{-\sqrt n}

2) If i prove bn b_n is convergent, then i also assume an a_n is convergent by direct comparison. Would i need to write anything else ?

Thanks in advance
(edited 11 years ago)
Original post by bijesh12
Determine the convergence/divergence of the series from n=2 to infinity
of :

(ln⁡n)e−n (\ln n)e^{-\sqrt n}

i know i can use the direct comparison test, ive done so by letting a function bn = 1/ ln (n), my questions are:

1) does it matter what function you use for b(n), as long as they fit the criteria that 0 < an < bn, and if so is it alright that i've used bn = 1/ ln(n).

2) If i prove bn is convergent, then i also assume an is convergent by direct comparison. I would i need to write anything else.

Thanks in advance


1n<1ln⁡n  ⇒  ∑n≥21n≤∑n≥21ln⁡n\dfrac{1}{n}<\dfrac{1}{\ln n}\;\Rightarrow\; \displaystyle\sum_{n\geq 2}\dfrac{1}{n}\leq\sum_{n\geq 2}\dfrac{1}{\ln n}

You need to find something else! :wink:
Reply 2
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bijesh12
OP
13
Original post by Lord of the Flies
1n<1ln⁡n  ⇒  ∑n≥21n≤∑n≥21ln⁡n\dfrac{1}{n}<\dfrac{1}{\ln n}\;\Rightarrow\; \displaystyle\sum_{n\geq 2}\dfrac{1}{n}\leq\sum_{n\geq 2}\dfrac{1}{\ln n}

You need to find something else! :wink:



i know im an idiot for choosing that one. But would this work as bn=ne−n b_n = ne^{-\sqrt{n}} , because it does converge, and it's true that an<bn a_n < b_n for all n≥1 n \geq 1
(edited 11 years ago)
Original post by bijesh12
i know im an idiot for choosing that one. But would this work as bn=n⋅e−n b_n = n\cdot e^{-\sqrt{n}} , because it does converge, and it's true that an<bn,    ∀n≥1 a_n < b_n,\;\; \forall n\geq 1


That does work. Bear in mind that you need to prove that n⋅e−nn\cdot e^{-\sqrt{n}} converges. Using bn=n⋅e−nb_n=\sqrt{n}\cdot e^{-\sqrt{n}} would be nicer to work with!
(edited 11 years ago)
Reply 4
Avatar for bijesh12
bijesh12
OP
13
Original post by Lord of the Flies
That does work. Bear in mind that you need to prove that n⋅e−nn\cdot e^{-\sqrt{n}} converges. Using bn=n⋅e−nb_n=\sqrt{n}\cdot e^{-\sqrt{n}} would be nicer to work with!


Thanks, but i would still need to prove that the latter function (the one you've suggested) is convergent again through the direct comparison, as i've tried both root and ratio tests and both are inconclusive :mad:,
Original post by bijesh12
Thanks, but i would still need to prove that the latter function (the one you've suggested) is convergent again through the direct comparison, as i've tried both root and ratio tests and both are inconclusive :mad:,


Hm, can I ask why? I'm assuming the exercise is asking you to prove that ana_n converges by comparison, not to use comparison for everything! :biggrin:

I made that suggestion specifically with the integral test in mind.
Reply 6
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bijesh12
OP
13
Original post by Lord of the Flies
Hm, can I ask why? I'm assuming the exercise is asking you to prove that ana_n converges by comparison, not to use comparison for everything! :biggrin:

I made that suggestion specifically with the integral test in mind.


The question is just asking to determine whether the series from n=2 till infinity of an=ln(n).e−n a_n = ln (n).e^{-\sqrt n} is convergent or divergent. It dosen't specify a method.

So it would be fine if i just that b_n is greater than a_n, and that it converges so by direct comparison so does a_n
(edited 11 years ago)
Original post by bijesh12
The question is just asking to determine whether the series from n=2 till infinity of an=ln⁡n⋅e−n a_n = \ln n\cdot e^-{\sqrt{n}} is convergent or divergent


Why the fuss about comparison then? Simply go: ∑n≥2ln⁡n⋅e−n≤∑n≥2n⋅e−n\displaystyle\sum_{n\geq 2}\ln n\cdot e^{-\sqrt{n}}\leq \sum_{n\geq 2}\sqrt{n}\cdot e^{-\sqrt{n}}
∑n≥2n⋅e−n\displaystyle\sum_{n\geq 2}\sqrt{n}\cdot e^{-\sqrt{n}} converges if and only if ∫2∞x⋅e−x  dx\displaystyle\int_2^{\infty} \sqrt{x}\cdot e^{-\sqrt{x}}\;dx converges. I'll let you finish off!
Reply 8
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bijesh12
OP
13
Original post by Lord of the Flies
Why the fuss about comparison then? Simply go: ∑n≥2ln⁡n⋅e−n≤∑n≥2n⋅e−n\displaystyle\sum_{n\geq 2}\ln n\cdot e^{-\sqrt{n}}\leq \sum_{n\geq 2}\sqrt{n}\cdot e^{-\sqrt{n}}
∑n≥2n⋅e−n\displaystyle\sum_{n\geq 2}\sqrt{n}\cdot e^{-\sqrt{n}} converges if and only if ∫2∞x⋅e−x  dx\displaystyle\int_2^{\infty} \sqrt{x}\cdot e^{-\sqrt{x}}\;dx converges. I'll let you finish off!


Sorry just checking, and thanks for all your help :smile:
Reply 9
Avatar for davros
davros
16
Original post by bijesh12
Determine the convergence/divergence of the series from n=2 to infinity
of :

(ln⁡n)e−n (\ln n)e^{-\sqrt n}

i know i can use the direct comparison test, ive done so by letting a function bn=n.e−n b_n = n.e^{-\sqrt n} , my questions are:

1) does it matter what function you use for b(n), as long as they fit the criteria that 0≤an≤bn 0 \leq a_n \leq b_n , and if so is it alright that i've used bn=n.e−n b_n = n.e^{-\sqrt n}

2) If i prove bn b_n is convergent, then i also assume an a_n is convergent by direct comparison. Would i need to write anything else ?

Thanks in advance



Original post by Lord of the Flies
Why the fuss about comparison then? Simply go: ∑n≥2ln⁡n⋅e−n≤∑n≥2n⋅e−n\displaystyle\sum_{n\geq 2}\ln n\cdot e^{-\sqrt{n}}\leq \sum_{n\geq 2}\sqrt{n}\cdot e^{-\sqrt{n}}
∑n≥2n⋅e−n\displaystyle\sum_{n\geq 2}\sqrt{n}\cdot e^{-\sqrt{n}} converges if and only if ∫2∞x⋅e−x  dx\displaystyle\int_2^{\infty} \sqrt{x}\cdot e^{-\sqrt{x}}\;dx converges. I'll let you finish off!


There's no need for integral tests, root tests or any other complicated method for this one - the point is that ene^n is so big that it can smash anything that is divided by it.

I would say something like: ln n < n so an<ne−na_n < ne^{-\sqrt{n}}

Also if x is positive, exe^x is bigger than any term in its expansion, so pick a nice juicy term with an even exponent e.g. x6/6!x^6/6!

Then
ex>x6/6!e^x > x^6/6!
so
en>n3/6!e^{\sqrt{n}} > n^3/6!
so
ne−n<n(n3/6!)=6!n2ne^{-\sqrt{n}} < \frac{n}{(n^3/6!)} = \frac{6!}{n^2}

and you can do a direct comparison with ∑1n2\sum \frac{1}{n^2}
Original post by davros
...


Thanks for bringing some mathematical sensibility to this thread! :wink:
Reply 11
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bijesh12
OP
13
Original post by davros
There's no need for integral tests, root tests or any other complicated method for this one - the point is that ene^n is so big that it can smash anything that is divided by it.

I would say something like: ln n < n so an<ne−na_n < ne^{-\sqrt{n}}

Also if x is positive, exe^x is bigger than any term in its expansion, so pick a nice juicy term with an even exponent e.g. x6/6!x^6/6!

Then
ex>x6/6!e^x > x^6/6!
so
en>n3/6!e^{\sqrt{n}} > n^3/6!
so
ne−n<n(n3/6!)=6!n2ne^{-\sqrt{n}} < \frac{n}{(n^3/6!)} = \frac{6!}{n^2}

and you can do a direct comparison with ∑1n2\sum \frac{1}{n^2}


Thanks, now I need help on part b) of the same question.

Assuming the series an=lnn.e−n a_n = ln n.e^{-\sqrt n} is convergent, determine whether ∑n=1∞an<∞ \sum^{\infty}_{n=1}{\sqrt{a_n} } < \infty

Firstly I would prove an\sqrt{a_n} is convergent.
(edited 11 years ago)
Original post by bijesh12
Thanks, now I need help on part b) of the same question.

Assuming the series an=lnn.e−n a_n = ln n.e^{-\sqrt n} is convergent, determine whether ∑n=1∞an<∞ \sum^{\infty}_{n=1}{\sqrt{a_n} } < \infty

Firstly I would prove an\sqrt{a_n} is convergent.


Use davros' method, but replace x66!\dfrac{x^6}{6!} with x1010!\dfrac{x^{10}}{10!}

(sorry for not making a better suggestion earlier by the way)
Reply 13
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davros
16
Original post by Lord of the Flies
Thanks for bringing some mathematical sensibility to this thread! :wink:


No worries!

I'd have to dig out the text books for a series requiring a really sensitive test, but when you get a negative exponential in there it can pretty much "kill off" anything multiplying it, so even making the series 'bigger' by replacing ln n with n doesn't damage the convergence :smile:
Reply 14
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bijesh12
OP
13
Thanks for all your help ! I got it all right in the end (Y)

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