The comparison test basically tells us that if there are two series #S_1# and #S_2# and we have that #S_1 <= S_2#, then if
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#S_2# converges, so does #S_1#
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#S_1# diverges, so does #S_2#
The trick to making the problem simpler is to choose #S_2# and #S_1# in such a way that it makes the algebraic computation fairly easy for you. Whenever you have to use the comparison test, try to look for geometric series which are similar to what is given in the question.
For this case, we know that:
#5/(2+3^n) < 1/2^n # because we are just multiplying the LHS by '5' and as n approaches infinity, the +2 in the denominator hardly makes a difference. However, on the RHS, we are taking #2^n#, which is much smaller than #3^n# as n gets infinitely large. Hence, the above inequality holds. (If you are not convinced, I encourage you to sketch a graph using a graphic calculator)
Now this problem is much simpler for us. We already know that #u_n = (1/2^n)# is a geometric series that converges to a finite sum #S# given by
#S = u_1/(1-r)# where r is the common ratio
Hence, for the above case, #u_1= 1# (because we replace 0 in u_n) and #r = 1/2#
Hence: #S = 1/(1-1/2) = 2#
And since #5/(2+3^n) < 1/2^n #, and #S# converges, by the Comparison test, the sum given in the question also converges (to a finite sum that is less than 2).
Hope this helps!
