Divergent and convergent series

Question

Determine whether the series listed below are divergent, absolutely convergent (hence convergent), or conditionally convergent. Indicate the tests or result applied to support

a. ∑ (to ∞) [(-1)^(n-1)e^(1/n)] / n

n=1

b. ∑ (to ∞) 1 / (n + n tan^(-1) n)

n=1

c. ∑ (to ∞) n! / (ln(n))

n=1

d. ∑ (to ∞) [(-n)^(3) + 1] / 7^n

n=1

Answer 1

(Note: I use lim as shorthand for the limit as n goes to infinity)

a) We perform the Alternating Series Test with a_n = e^(1/n)/n. The Alternating Series Test says that for the series ∑(-1)^(n-1) a_n converges if 0 < a_{n+1} < a_n for all n in N, and lim a_n=0. Note that a_{n+1} = e^{1/(n+1)}/(n+1). This is positive for all n in N. Also, remember that a bigger denominator means a smaller number. This means 1/(n+1) < 1/n for all n in N, so then 0 < e^{1/(n+1)}/(n+1) < e^{1/n}/n since e^x is an increasing function. Therefore 0 < a_{n+1} < a_n holds. Now we find the limit of a_n. Let L = lim a_n.

L = lim e^{1/n}/n

ln L = ln (lim e^{1/n}/n)

ln L = lim [ln (e^(1/n)) - ln n] (log rules, limit of ln is ln of limit)

ln L = lim ((1/n)ln e) - lim ln n (breaking up limit, log rules)

ln L = 0 - ∞ = -∞

e^(ln L) = e^(-∞)

L = 0.

Both conditions are satisfied so the series converges by Alternating Series Test. However, we must test if ∑ |(-1)^(n-1)e^(1/n)/n | = ∑ e^{1/n}/n in order to determine whether our original series in part (a) converges *absolutely*. Since e^(1/n) ≥ 1 for all n in N, e^{1/n}/n ≥ 1/n. 1/n is the harmonic series and is known to be divergent, meaning e^{1/n}/n diverges by the Direct Comparison Test. Therefore, our series converges conditionally, NOT absolutely.

b) We’ll do the limit comparison test with the known divergent series 1/n since it looks pretty similar to our series.

lim (1/n)/(1/(n + narctann))

= lim (n+narctann)/n (this has indeterminate form ∞/∞ so this calls for L’Hôpital)

= lim (1+ arctann + [n/(1+n^2)])=1+ π/2 > 0

Therefore the series diverges by limit comparison test.

c) Note that 1 ≤ n! for all n in N, so 1/ln n ≤ n!/ln n for all n in N. ∑1/ln n is known to diverge, so ∑n!/ln n diverges by direct comparison test.

d) We can break this series into two series and see if each part is convergent:

∑((-n)^3+1)/7^n = ∑[((-n)^3)/(7^n) + 1/(7^n)] = ∑((-n)^3)/(7^n) + ∑(1/7)^n.

Since |1/7|<1, ∑(1/7)^n is a convergent geometric series and converges absolutely. It remains to show ∑((-n)^3)/(7^n) converges. We’ll use the ratio test. Let L = lim ⁿ√|((-n)^3)/(7^n)|. Then

L = lim ⁿ√[(n^3)/(7^n)] (made everything under absolute value positive)

L = lim n^(3/n)/7 (thinking of nth root as fractional exponent)

L = (1/7) lim n^(3/n)

7L = lim n^(3/n)

ln(7L) = ln (lim n^(3/n))

ln(7L) = lim (ln n^(3/n))

ln(7L) = lim (3 ln n)/n (log rules, multiply (3/n)*ln n)

ln(7L) = lim (3/n)/1 (L’Hôpital: we had ∞/∞ in the last step)

ln(7L) = 0

e^ln(7L) = e^0

7L = 1

L = 1/7 < 1

So then ∑((-n)^3)/(7^n) converges absolutely by the Ratio test. Since both parts of our original series converge absolutely, the series all put together converges absolutely.

Hope this helps! :)