I really need help on the remainder of my HW assignment. Any help is appreciated, so post up if you think you may have something useful to help find the answer (if you are not sure the answer/work you provided may not be correct, please state so).

Also, if you scratch my back, I'll scratch yours

Part 1:
7. Let f(x)= 1 / [x ((lnx )^2)].

A) Find the domain of f.

B.) Find intergral f(x)dx.











C) Evaluate integral f(x)dx from 2 to infinity.






D) Use part c) to determine whether the series ∑(2 to infinity)f(n) converges or
diverges. What test have you applied to make that determination?







E) How many terms in the series ∑(2 to infinity)f(n) must be added to estimate this
sum to within 0.01?





Part 2:
Determine whether each series converges or diverges. State the test you have
applied in making that determination. Find the exact sum of each convergent
series whenever possible.

1) ∑(k=2 to infinity)(e/3)^k


2) ∑(n=1 to infinity)[(square root n) / (2n^2 - 1)]


3) ∑(j=1 to infinity)arctanj

1)
a) domain of f is all Reals except x=0 and x=1 since the function is discontinuous at those points
the integral can be found by using the u-substitution method; specfically let u = ln(x) and notice then that du = (1/x) *dx

so the integral becomes int[ (1/u^2)*du ]
the indefinite integral of 1/u^2 is simply -1/u +C
now simply substitute u back in to get the answer: (-1/lnx) + C

c) to take the improper integral from 2 to infinity just pretend you're integrating from 2 to some value y. [the premise here is that you will be taking the limit of this integral as y -> infinty]

so, we found the indefinite integral in part (b), we just need to plug in 2 and y to get: (-1/ln y) - (-1/ln(2) ) to be the integral from 2 to y of the function
next, we take the limit of this as y -> infinity, and we get: 0 + 1/ ln(2) = 1/ln(2) as our final answer

d) we can use a direct comparison test to {1/n} from 2 to infinity:

0 <= 1/[n*ln(n)^2] <= 1/n

since 1/n converges to zero as n approaches infinity, by comparison, f(n) = 1/ [n*ln(n)^2] must also converge

Part 2:

1) (e/3)^k can be rewritten as 1/ (3/e)^k. Using the P-series test for convergence (letting n = 3/e), we can see that this series must converge since p = k>=2 > 1 ~8.743 is the sum of the convergent series
2) by the comparison test, the series converges and the exact sum of the series is ~1.841
3) the series diverges by the limit comparison test

This post was edited by JDota72 on Nov 7 2014 07:27pm

Thanks JDota72, I will check over your posts tomorrow.

If I need additional help I will post up or pm JDota72

Just a few errors of what JDota72 has stated.

1. The Domain of your function is {x ∈ R | 0<x<1 or x>1} because ln is not define for negative numbers

2. For question d) JDota72 is correct to say that your series converges, however his method is incorrect because the series 1/n is actually DIVERGENT.
Instead I would say that your series is convergent using the integral test, which states that is f is a continuous, positive decreasing function then your series converges iff the improper integral converges. (your professor will probably want you to show f is positive and decreasing. can easily be done taking the first derivative)

This post was edited by cdexswzaq on Nov 7 2014 08:24pm

oh yes, he is right

Currently need assistance with:

E) How many terms in the series ∑(2 to infinity)f(n) must be added to estimate this
sum to within 0.01?

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I need to look thru my book more to complete #2 and #3 in Part 2. JDota gave the following tips:

Part 2:
2) by the comparison test, the series converges and the exact sum of the series is ~1.841
3) the series diverges by the limit comparison test

So if anyone has more detail on these two, that would be great!

This post was edited by CamelFinger on Nov 9 2014 05:48pm

Use the error estimate from the Integral Test
∫ (from k to ∞) 1dx / [x ((lnx )^2)] ≤ 0.01
(-1/lnx)(from k to ∞) ≤ 0.01

1/lnk ≤ 0.01
lnk ≥ 100
k ≥ e^100

so you would need to add k ≥ e^100 terms

for 2)
Σ√n/(2n^2-1)
we know 1/n^(3/2) is a convergent p series
so we can use the limit comparison test
= lim as n->∞ [√n/(2n^2-1)] / [1/n^(3/2)]
= lim as n->∞ n^2/(2n^2-1)
= 1/2 > 0
therefore since 1/n^(3/2) converges √n/(2n^2-1) also converges


for 3)
using the divergence test which says
if ∑(1 to ∞, a(n) ) converges, then lim (n -> ∞, a(n)) = 0.

Therefore the contrapositive of this statement would be

If lim (n -> ∞, a(n)) is NOT equal to zero, then ∑(1 to ∞, a(n) ) diverges.

lets evaluate the lim as n -> ∞ of arctan(n)

lim arctan(n)
n -> ∞

the limit is pi/2. (you can check this yourself)

and therefore since the limit is not 0, the series is divergent

This post was edited by cdexswzaq on Nov 9 2014 08:41pm

cdex u are awesome!

Thanks for the help JDota and cdex!

/Thread