How do i show this series converges?

Question

Show that this series converges to ln(2). Consider the n-th Taylor Polynomial for ln(x) at x=1 and the corresponding term

∞
∑ ((-1)^(n+1))*(1/n)
n=0

I have absolutely no idea how to do this, can anyone help me plz!?!?!?!

Answer 1

The convergence is a simple consequence of the Alternating Series Test: the series alternates and the terms monotonically decrease to 0.

As for what it converges to, do as the problem suggests, use the power series for
ln(1+x)=x-(1/2)x^2+(1/3)x^3...
=sum (-1)^(n+1) x^n /n.

This series converges to ln(1+x) for -1

Answer 2

question 2a: (i presumed you recommend (-2)^n - a million/3^n. in case you meant (-2)^(n+a million) / 3^n, see the different answer by using Rafaele) you will tutor that an limitless sum in straightforward terms converges if the lim n -> infinity of {An} = 0. So, lim n -> infinity of {An} = lim n -> infinity (-2)^n - lim n -> infinity a million/3^n = lim n -> infinity (-2)^n (which would not exist. as a result, by using contraposition, this sequence would not converge. question 2c: First tutor that {e^(a million/n)} is Cauchy and/or converges. Then, {-e^(a million/n)} is likewise convergent (bounded and has comparable decrease as n -> infinity, i.e. 0). because of the fact the sequence is truly convergent, the order of the sum would not remember (it is going to continuously get the comparable fee). For all advantageous epsilon there exists n such that |An+a million| + ... < epsilon. elect this epsilon. Rearrange the finite sum as much as n so as that each and every of the words cancel, different than e^a million and - e^(a million/(n+a million)). considering that e^(a million/n) is Cauchy, the limitless sum is decrease than epsilon + e - e^(a million/(n+a million)) < epsilon + e - a million. yet it is the definition of the decrease as n techniques infinity. So, the limitless sum is comparable to e - a million. question 3: merely be conscious the ratio try. no could tutor this (i think of). lim n-> infinity of [ (x-3)^(n+a million) / 5^(n+a million) ] / [ (x-3)^n / 5^n] = lim n-> infinity of (x-3) / 5 To converge, this decrease could be decrease than one. So, x < 5 + 3 = 8. additionally, it may well be better than -a million. So, x < -5 + 3 = -2. So, -2 < x < 8. be conscious that the obstacles x = -2 and x = 8 additionally must be checked. considering that this could be a geometrical sequence, we state that the sequence is divergent at the two places. whilst convergent, the sum is a / (a million-r), the place a = a million and r = (x-3)/5. So, limitless sum = a million / (a million - (x-3)/5) = 5 / (8-x).

Answer 3

∞
∑ ((-1)^(n+1))*(1/n) = -1+1-1/2+1/3-1/4+1/5-....
n=0

To show a series converges we need to know the individual terms convert to zero, which they do. But that is not sufficient.

∞
∑ ((-1)^(n+1))*(1/n) = (-1+1)+(-1/2+1/3)+(-1/4+1/5)+
n=0
(-1/6+1/7)....

=0 -1/6 -1/20 -1/42 -1/72-...

Factoring the denominators:

=0 -1/(2*3) -1/(4*5) -1/(6*7)-1/(8*9)

Noting these factors. The terms have absolute values smaller than:
∞
∑ (1/n^2) = 1+1/4+1/9+1/16, which converges
n=1
So our series must converges.

Hope this helps. The key is to show that the series has terms closer to zero than a know convergent series.

Answer 4

call a_n= 1/n a_n+1= 1/(n+1) so the alternate serie is convergent by Leibnitz criteria