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Can anyone see how to solve this series

  • Thread starter Dell
  • Start date
  • #1
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i need to say prove that the following converges with n from 1 to infinity

Σ((-1)n-1)*((lnp(n)))/n) (p>0)

first thing i see here is integration, so i can tell if the series of absolute values converges or not, if it does then i know that my original series must converge.

t=ln(n)
dt=dn/n

so i have an integral from 0 to infinity of

tp dt

which is a simple integration and i find that the absolute values' series diverges which tells me that my series may conditionally converge if
lim an =0 (which it is)
and an>a(n+1) which it is not since the series keeps on rising in its absolute value, because of "p".

so how do i prove that the series converges?
 

Answers and Replies

  • #2
CompuChip
Science Advisor
Homework Helper
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1) Note that your integral runs from 1 to infinity, not from 0
2) The convergence of the integral depends on the value of p
3) If I read your post correctly you think you have shown that it always diverges, so why are you asking for a proof that it converges?
 
  • #3
590
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1) the original integral(dx) would have been from 1 to infinity, but after substitution(t=lnx) it runs from ln(1) which is 0
2)originally i also thought that the convergence depends on p, but it is given that p>0, and for any p, (lnx)^p/x is smaller than 1, not sure that it depends on p, plus the answer in the book is independant of p
3) i have shown that Σ|An| diverges, but that does not mean that ΣAn diverges necessarily, that is sort of the whole question, how do i find if it does or doesnt
 
  • #4
CompuChip
Science Advisor
Homework Helper
4,302
47
Right, I'm sorry, I completely missed your point apparently...

So you have shown that there is no absolute convergence for any p, the question is then if there is conditional convergence, right?
I think the Leibniz test may help you with that.
 
  • #5
590
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i also thought that, only that i have no way of proving that an>an+1, in fact i think that an is not necessarily bigger than an+1,
for small n's, an is even smaller than an+1, for example a(1)=0 a(2)>0 because of ln,

is it enough to say that for n->inf, an>an+1 and so the seires conditionally converges, for smaller n's it really depends on p, but for very large, infinite, n- ln(n)^p is always smaller than n,
 

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