How to determine the convergence or divergence of series? q1. sigma(n=1, infinity) #(-1)^(n+1)/n# q2. sigma(n=2, infinity) #(-1)^n/ln(n)# q3. sigma(n=0, infinity) #(-1)^n/(n!)# q4. sigma(n=1, infinity) #(-1)^n/sqrt(n)#

1 Answer
VNVDVI
Apr 13, 2018

Using the Alternating Series Test, all converge.

Explanation:

Q1. This is an alternating series in the form #sum_(n=1)^oo(-1)^(n+1)b_n# where #b_n=1/n#.

The Alternating Series Test tells us that if #b_n# is decreasing (ultimately) and has positive terms and #lim_(n->oo)b_n=0,# the alternating series #sum_(n=1)^oo(-1)^(n+1)b_n# converges.

Here, #lim_(n->oo)1/n=0, 1/n# is decreasing because as #n# gets bigger, the denominator gets bigger, causing #b_n# to always decrease.

Positivity is not a concern, as we start at #n=1# and therefore have no negative terms for #b_n.#

Thus, we have convergence.

Q2. This is yet another alternating series, this time in the form #sum_(n=2)^oo(-1)^nb_n, b_n=1/ln(n)#.

Here, we should note that the specific exponent on the #-1# does not change the fact that the series is alternating, so long as the exponent is in the form #n+-a, a# is some integer.

On #[2, oo), 1/ln(n)# is positive as only logarithms between #(0,1)# are negative. It is decreasing -- the denominator increases causing the overall sequence to decrease.

Finally, #lim_(n->oo)1/ln(n)=0#.

Once again, convergence by the Alternating Series Test.

Q3. This is alternating. #b_n=1/(n!)#. We will have to prove it is decreasing, or, #b_n>=b_(n+1)#:

#1/(n!)>=1/((n+1)!)#

Recall that #(n+1)! = (n+1)n!# -- we can strip terms out of factorials.

#1/(cancel(n!))>=1/((n+1)cancel(n!))#

#1>=1/(n+1)# for all #n# on #[0, oo)# is true.

Now, for the limit, we cannot evaluate right away, so we take the following steps:

#0<=1/(n!)<=1/n#

#lim_(n->oo)0=0#

#lim_(n->oo)1/n=0#

Then, by the Squeeze Theorem

#lim_(n->oo)1/(n!)=0#

Convergent.

Q4. This is again alternating, with #b_n=1/sqrtn.#

#b_n# is positive on #[1,oo)# and decreasing; the growing denominator entails a decreasing sequence.

#lim_(n->oo)1/sqrtn=0#

Convergent.

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