How do you determine if #a_n=6+19+3+4/25+8/125+16/625+...+(2/5)^n+...# converge and find the sums when they exist?

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Answer 1 · 2017-10-22T18:40:08

See below.

From the fourth term on we have a geometric series.

The sum of this can be found using:

#a((1-r^n)/(1-r))#

Where #a# is the first term, #r# is the common ratio and #n# is the nth term.

So from 4th term to infinity we have:

#4/25((1-(2/5)^n)/(1-(2/5)))= 1/(3/5)=5/3*4/25=20/75=4/15#

#sum_(n=2)^(oo)(2/5)^n=4/15# ( Convergent )

Then we have:

#sum(6+19+3)= 28#

And:

#sum(6+19+3) + sum_(n=2)^(oo)(2/5)^n=424/15# ( Convergent )