Question #213799

A geometric progression has first term and common ratio

i)                  Find the set of values of x for which geometric progression has a sum to infinity.

  • ii)                Find the value of x for which the sum to infinity of geometric progression 
1
Expert's answer
2021-07-06T14:32:22-0400

I haven't got any information on first term and common ratio of the geometric progression. I assume that the first term is a0a\ne 0 and common ratio is x.

Then n-th term is axn1ax^{n-1} and sum of the first n term is

sn=a1++an=a+ax++axn1=axn1x1=a1xa1xxns_n=a_1+\dots+a_n=a+ax+\dots+ax^{n-1}=a\frac{x^n-1}{x-1}=\frac{a}{1-x}-\frac{a}{1-x}x^n (x1x\ne 1)

sns_n tends to a finite limit as n+n\to+\infty if and only if xnx^n tends to a finite limit as n+n\to+\infty, i.e., if and only if x<1|x|<1. With this condition we have limn+xn=0\lim\limits_{n\to+\infty}x^n=0 and n=1+axn1=limn+sn=a1x\sum\limits_{n=1}^{+\infty}ax^{n-1}=\lim\limits_{n\to+\infty}s_n=\frac{a}{1-x}.

In the case x=1 we have an=axn1=aa_n=ax^{n-1}=a and sn=ans_n=an diverges to infinity.

Answer. (i) x<1|x|<1; (ii) n=1+axn1=a1x\sum\limits_{n=1}^{+\infty}ax^{n-1}=\frac{a}{1-x}.


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