Convergent/Divergent series

Question

Are the series convergent or divergent? What is the sum of each convergent series?

a.) ∑ (going to ∞) e^(i) / 3^(1-2)

i=1

b.) ∑ (going to ∞) (4 / (n^(2) + 4n + 3)) + (3^(n-1) / 4^(n))

i=1

c.) ∑ ln (n / n+3)

i=1

d.) ∑ (going to ∞) ln(n) - ln(1+2n)

n=1

Answer 1

(Note: for brevity I will use lim to abbreviate the limit as n goes to infinity and ∑ to abbreviate the sum as n goes from 1 to infinity.)

a) (assuming 3^(1-2) is supposed to be 3^(i-2)) By exponent rules, we may rewrite 3^(n-2) as 3^n(3^(-2))=(1/9)3^n. So then ∑e^n/3^(n-2)=∑9e^n/3^n=9∑(e/3)^n. Since e≈2.7, e is certainly smaller than 3, meaning e/3<1. That means we have a convergent geometric series so we use the formula (first term)/(1-r) with r being the ratio. So then we get 9∑(e/3)^n=9((e/3)(1-e/3)). After a little simplification the answer is 9/(3-e).

b) We can break this one into the two series ∑4/(n^2+4n+3) + ∑3^(n-1)/4^n. Let’s check if each series is convergent, starting with ∑4/(n^2+4n+3). We can do the limit comparison test with ∑1/n^2 which is known to converge (and famously converges to (π^2)/6 which still blows my mind). We have that lim (1/n^2)/(4/(n^2+4n+3)) = lim (n^2+4n+3)/4n^2 = 1/4, and since 0<1/4<∞, the series converges. Now let’s try ∑3^(n-1)/4^n. We can rewrite this as (1/3)∑(3/4)^n using the same method as part (a), and since 3/4<1, the series is a convergent geometric series. So since both series converge, their sum must converge as well.

Now let’s try adding up ∑4/(n^2+4n+3). If we do partial fraction decomposition we can rewrite this as ∑(2/(n+1)-2/(n+3)). Now let S_n=∑k=1 to n (2/(k+1)-2/(k+3)) be the sequence of partial sums. We write

S_n = 2/2 - 2/4 + 2/3 - 2/5 + 2/4 - 2/6 + 2/5 - 2/7 + … + 2/(n-2) - 2/n + 2/(n-1) - 2/(n+1) + 2/n - 2/(n+2) + 2/(n+1) - 2/(n+3).

This is what’s called a telescoping sum because almost all the terms cancel out and we’re left with only 4 terms: S_n = 1 + 2/3 - 2/(n+2) - 2(n+3). Now we get lim S_n = 5/3, meaning ∑4/(n^2+4n+3)=5/3.

Now we’ll look at the second series. We have that ∑3^(n-1)/4^n = (1/3)∑(3/4)^n = (1/3)[(3/4)/(1-3/4)] = 1. Therefore ∑4/(n^2+4n+3) + ∑3^(n-1)/4^n = 5/3 + 1 = 7/3.

c) Note that ln(n/(n+3))=lnn - ln(n+3) by log rules. So we’ll get a telescoping sum again. Let S_n = ∑k=1 to n lnn - ln (n+3). Then S_n = ln1 - ln4 + ln2 - ln5 + ln3 - ln6 + ln4 - ln7 + ln5 - ln8 + … + ln(n-3) - lnn + ln(n-2) - ln(n+1) + ln(n-1) - ln(n+2) + lnn - ln(n+3) = ln2 + ln3 - ln(n+2) - ln(n+3) after doing the cancellations and realizing ln1 = 0. We can simplify this to S_n = ln(6/[(n+2)(n+3)]). Now lim S_n = lim ln(6/[(n+2)(n+3)]) = ln (lim 6/[(n+2)(n+3)]) = ln0 = -∞, so the series diverges.

d) We can simplify this to ∑ln(n/(1+2n)). Using the nth term test, we get lim ln(n/(1+2n)) = ln (lim n/(1+2n)) = ln (1/2) which is not 0, so the series diverges.

Hope this helps!!

Answer 2

It is not my job to DO your homework for you, i.e. you should work on all of these problems yourself.

I will give you some hints.

They all converge.

a) is a slightly disguised geometric series

b) is the sum of 2 convergent series: in the first series write the fraction as the difference of 2 fractions, 1/(n+1) and 1/(n+3)...which telescopes and the second series which is geometric

c) write the logarithm of the fraction as a sum of logarithms...then it will telescope

d) this one telescopes directly (write out the first few terms!