#sum_(n=1)^inftyn#is divergent; why?#sum_(n=1)^infty(-1)^n#is divergent; why?#sum_(n=1)^infty(1)/(3^(n-1))#is convergent #S=3/2#; why?#sum_(n=1)^infty(1)/(3^n)#is convergent #S=2/9#; why?#sum_(n=1)^infty(4n^2-n^3)/(10+2n^3)#is divergent; why?#sum_(n=1)^infty(2/3^n+2/(3n))#is divergent; why?#sum_(n=1)^infty9^(2-n)4^(n+1)#is convergent #S=259.2#; why?

I would mainly like to understand the best method of trial and error convergence/divergence in every of this problems. I don"t desire to garbage time ~ above a work-intensive process during a test...

You are watching: Determine whether the series converges or diverges.

(Please don"t teach shortcuts unless they room proven come work!)

Calculus Andrea S.
Apr 28, 2017

(2) is not an exact and (4) is incorrect

Explanation:

(1) The collection does not respect Cauchy"s necessary problem since:

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

then it cannot converge.

In reality the partial amount #s_N# is:

#s_N = sum_(n=1)^N n = 1+2+3+...+N = (N(N-1))/2#

so:

#lim_(N->oo) s_N = oo#

(2) Again, the series does not respect Cauchy"s necessary problem since:

#lim_(n->oo) (-1)^n#

does not exist.

In truth if #s_N# is the partial sum we can see that:

#s_1 = -1#

#s_2 = 0#

and we can prove by induction that:

#s_N = (-1 " for " N " odd "),(0 " for " N " also "):#

so that #lim_(N->oo) s_N# does not exist ( and also then the collection is undetermined, no divergent).

See more: Wanderlei Silva Vs Chuck Liddell Vs Wanderlei Silva Full Fight Result

(3) This can be diminished to the amount of a geometric series, and also we understand that:

#sum_(n=0)^oo a^n = 1/(1-a)# because that #abs a

In truth we have:

#sum_(n=1)^oo 1/3^(n-1)#

substitute #k= n-1#

#sum_(n=1)^oo 1/3^(n-1) = sum_(k=0)^oo 1/3^k = sum_(k=0)^oo (1/3)^k = 1/(1-1/3) = 3/2#

(4) together above:

#sum_(n=1)^oo 1/3^n#

is simply a geometric series lacking the very first term: if we add and subtract #1# and note that #(1/3)^0 = 1# us get:

#sum_(n=1)^oo 1/3^n = -1 + 1 + sum_(n=1)^oo 1/3^n = -1 + sum_(n=0)^oo (1/3)^n = -1+1/(1-1/3) = -1+3/2 = 1/2#

You can likewise solve it note that:

#sum_(n=1)^oo 1/3^n = 1/3 sum_(n=1)^oo 1/3^(n-1)#

so indigenous the vault exercise:

#sum_(n=1)^oo 1/3^n = 1/3 sum_(n=1)^oo 1/3^(n-1) = 1/3 xx 3/2 = 1/2#

(5) we have:

#sum_(n=1)^oo (4n^2-n^3)/(10+2n^3)#

Also this collection does not meet Cauchy"s necessary problem as:

#lim_(n->oo) (4n^2-n^3)/(10+2n^3) = -1/2#

In reality if we division numerator and denominator by #n^3#:

#a_n = (4/n-1)/(10/n^3+2 ) #

For #n > 4# we have actually that #a_n# is an adverse so if us decrease the denominator we have actually a negative quantity the is larger in absolute value, the is:

#a_n = (4/n-1)/(10/n^3+2 ) for #n > 4#

Now because that #n > 8# we have:

#4/n

so:

#1-4/n > 1-1/2#

#1-4/n > 1/2# because that #n > 8#

and then:

#a_n because that #n > 8#

Now consider the partial sum for #N :

#s_N = s_7 + sum_(n=8)^N a_n

and because #lim_(N->oo) s_7 -(N-8)/4 = -oo#

then also:

#lim_(N->oo) s_N = -oo#

(6) us have:

#sum_(n=1)^oo (2/3^n+2/(3n)) = 2/3 sum_(n=1)^oo (1/3^(n-1)+1/n)#

but then:

#a_n = 1/3^(n-1)+1/n > 1/n#

and as the harmonic series is divergent then by comparison likewise this collection is divergent.

(7) Again a geometric series:

#sum_(n=1)^oo 9^(2−n)4^(n+1) = 4*4*9 sum_(n=1)^oo 4^(n-1)/9^(n-1) = 144 sum_(n=1)^oo (4/9)^(n-1) = 144*1/(1-4/9) =9/5*144 = 259.2#