#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...

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Calculus
1 answer
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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).

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(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#