\sum_(n=1 )^(\infty ) ((-1)^(n)ln(n))/(n)

Question

(a) Determine whether the series is absolutely convergent, conditionally convergent, or divergent.

b) Summarize your work and conclusions from part (a) in 5–7 complete sentences. Make sure to explicitly mention any tests used

(c) Approximate the sum of the series using the 6th partial sum, s6. Round to four decimal places.

(d) Estimate the error involved in using s6 to approximate the sum of the series (rounded to 4 decimal places). Do you consider s6 to be a good approximation? Explain your reasoning.

(e) Find the first 8 partial sums of the series rounded to four decimal places, and then plot them

f)What is your rough guess for the sum of the series?

Answer 1

Test for absolute convergence:

∑_{(n=1 to ∞)} l (-1)ⁿ ln(n)/n l = ∑_{(n=1 to ∞)} [ln(n) / n] = 0 + ln2/2 + ln3/3 + ...

For n > 1, ln(n)/n > 0 and the terms are decreasing. So, the integral test is applicable.

∫_{(x=2 to ∞)} [ln(x)/x]dx = lim_b→∞ ∫_{(2 to b)} [ln(x)/x]dx =

lim_b→∞ [(1/2)(lnx)²]_{(2 to b)} = (1/2)lim_b→∞[(lnb)² - (ln2)²] = ∞

By the Integral Test, the series diverges. So, the given series does NOT converge absolutely.

By the Alternating Series Test, the given series converges.

So, the given series converges conditionally.

Answer 2

You need 2 pieces of information to complete this problem.

This is a Leibnitz series, i.e. alternating signs and decreasing terms...therefore it converges.

The error committed by stopping at term n does not exceed the n+1 term.

Now you just have some arithmetic to do...and I will not do that for you.

Also I would suggest you use a plotting program, e.g. Desmos.

Be careful: this series apparently converges very slowly!