How do you test the series #sum_(n=0)^(oo) n/((n+1)(n+2))# for convergence?

Question

Christopher Allers · Accepted Answer

undefined To test the series $ \sum_{n=0}^{\infty} \frac{n}{(n+1)(n+2)} $ for convergence, we can use the ratio test.

1. Calculate $ \lim_{n \to \infty} \frac{a_{n+1}}{a_n} $, where $ a_n = \frac{n}{(n+1)(n+2)} $.

2. If the limit is less than 1, the series converges absolutely. If it's greater than 1 or the limit doesn't exist, the series diverges. If the limit equals 1, the ratio test is inconclusive.

Let's apply the ratio test:

$$ \lim_{n \to \infty} \frac{\frac{n+1}{((n+1)+1)((n+1)+2)}}{\frac{n}{(n+1)(n+2)}} $$

$$ = \lim_{n \to \infty} \frac{n+1}{(n+2)(n+3)} \times \frac{(n+1)(n+2)}{n} $$

$$ = \lim_{n \to \infty} \frac{n+1}{n} \times \frac{n+1}{n+3} $$

$$ = \lim_{n \to \infty} \frac{(n+1)^2}{n(n+3)} $$

$$ = \lim_{n \to \infty} \frac{n^2 + 2n + 1}{n^2 + 3n} $$

$$ = 1 $$

Since the limit equals 1, the ratio test is inconclusive. Therefore, we need to use another test, such as the divergence test or comparison test, to determine the convergence or divergence of the series.

Charles Anctil · Answer

The series diverge

Perform the limit comparison test #a_n=n/((n+1)(n+2))# and #b_n=1/n#, this series diverge #a_n>0# and #b_n>0#, #AA n in NN# #lim_(n->oo)a_n/b_n=lim_(n->oo)((n/((n+1)(n+2)))/(1/n))# #=lim_(n->oo)(n^2/((n+1)(n+2)))# #=lim_(n->oo)(n^2/((n^2+3n+2)))# #=lim_(n->oo)(1/((1+3/n+2/n^2)))# #=1# We conclude that, by the limit comparison test that the series #a_n# diverge

Elijah Abram · Answer

The series:

#sum_(n=0)^oo n/((n+1)(n+2))#

is divergent.

The series has only positive terms, so we can use the limit comparison test to compare it with the harmonic series: #lim_(n->oo) (n/((n+1)(n+2)))/(1/n) = lim_(n->oo) n^2/(n^2+3n+2) = 1# As the limit is finite and positive the two series have the same character, and we know the harmonic series to be divergent, thus also the series: #sum_(n=0)^oo n/((n+1)(n+2))# is divergent.

Emily Ammerman · Answer

We can use the integral test to find it diverges.

Using the integral test, we find: #int x/((x+1)(x+2)) dx = int (2/(x+2) - 1/(x+1)) dx# #color(white)(int x/((x+1)(x+2)) dx) = 2 ln abs(x+2) - ln abs(x+1) + C# #color(white)(int x/((x+1)(x+2)) dx) = ln (abs(x+2)^2/abs(x+1)) + C# #color(white)(int x/((x+1)(x+2)) dx) > ln (abs(x+1)^2/abs(x+1)) + C = ln (abs(x+1)) + C# #color(white)(int x/((x+1)(x+2)) dx) -> oo" "# as #x -> oo# So: #sum_(n=0)^N n/((n+1)(n+2)) -> oo" "# as #N -> oo#

Christopher Agresta · Answer

See below.

#n/((n+1)(n+2))=2/(n+2)-1/(n+1)# then #sum_(n=0)^oo n/((n+1)(n+2)) = 2 sum_(n=0)^oo 1/(n+2) - sum_(n=0)^oo 1/(n+1)# Now considering #H = sum_(n=1)^n 1/n# we have #sum_(n=0)^oo n/((n+1)(n+2)) = 2(H-1)-H = H-2# but as we know #H# is the so called harmonic series https://tutor.hix.ai(mathematics) know as divergent. Resuming #sum_(n=0)^oo n/((n+1)(n+2))# is divergent.