How do you know if the series #sum 1/(n^(1+1/n))# converges or diverges for (n=1 , ∞) ?

Question

Emily Ammerman · Accepted Answer

The series diverges

To test the convergence of the series #sum_{n=1}^oo a_n#, where #a_n=1/n^(1+1/n)# we carry out the limit comparison test with another series #sum_{n=1}^oo b_n#, where #b_n=1/n#, We need to calculate the limit #L = lim_{n to oo }a_n/b_n = lim_{n to oo} n^{-1/n}# Now, #ln L = lim_{n to oo}( -1/n ln n) = 0 implies L=1# According to the limit comparison test , since this limit is a finite nonzero number, the series #sum_{n=1}^oo a_n# if and only if #sum_{n=1}^oo b_n# converges. However, it is well known that #sum_{n=1}^oo b_n# diverges, and hence our series diverges.

Charles Arocha · Answer

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

is divergent.

Using the properties of exponents: #1/n^(1+1/n) = 1/(n * n^(1/n)) = 1/n n^(-1/n)# Note now that: #n^(-1/n) = (e^(ln n))^(-1/n) = e^(-ln n /n)# As: #lim_(n->oo) ln/n = 0# and the exponential function #e^x# is continuous, then: #lim_(n->oo)e^(-ln n /n) = e^((lim_(n->oo) -ln/n )) = e^0 = 1# Consider now the harmonic series #sum_(n=1)^oo 1/n# that we know to be divergent. Using the limit comparison test: #lim_(n->oo) (1/n^(1+1/n))/(1/n) = lim_(n->oo) (1/n n^(-1/n))/(1/n) = lim_(n->oo) n^(-1/n) = 1# we can see that, as the limit of the ratio is finite, the two series have the same character and also: #sum_(n=1)^oo 1/n^(1+1/n)# is divergent.

Emilia Aguilar · Answer

undefined To determine if the series sum 1/(n^(1+1/n)) converges or diverges for n=1 to infinity, we can use the limit comparison test. We compare it with the series sum 1/n.

lim (n→∞) (1/(n^(1+1/n))) / (1/n)

Simplify the expression inside the limit:

lim (n→∞) n / (n^(1+1/n))

Now, rewrite n as n^(1/1) to match the denominator's form:

lim (n→∞) n^(1/1) / (n^(1+1/n))

Apply the quotient rule for exponents:

lim (n→∞) n^((1-1)/(1+1/n))

Simplify the exponent:

lim (n→∞) n^(1/(1+1/n))

As n approaches infinity, the exponent approaches 1/(1+0) = 1.

Therefore, the limit becomes:

lim (n→∞) n^1

Since the limit is infinity, the series 1/(n^(1+1/n)) diverges by the limit comparison test when compared to the harmonic series 1/n.